General and Middle Terms

In the binomial expansion for \((a+b)^n\), we observe that the
first term is \({ }^n \mathrm{C}_0 a^n\),
the second term is \({ }^n \mathrm{C}_1 a^{n-1} b\),
the third term is \({ }^n \mathrm{C}_2 a^{n-2} b^2\), and so on.
Looking at the pattern of the successive terms we can say that the \((r+1)^{t h}\) term is \({ }^n \mathrm{C_r}a^{n-r} b^r\).
The \((r+1)^{t h}\) term is also called the general term of the expansion \((a+b)^n\) and it is denoted by \(\mathrm{T}_{r+1}\).
Thus \(\mathrm{T}_{r+1}={ }^n \mathrm{C}_r a^{n-r} b^r\).

Example 1: Find the 4th term of the expansion \((3 x+2 y)^5\).

Solution: Step 1: Identify variables
To find a specific term using the general term formula \(\mathrm{T}_{r+1}={ }^n \mathrm{C}_r a^{n-r} b^r\), identify the relevant variables from the expression \((3 x+2 y)^5\).
\(n=5\) is the power, \(a=3 x\) is the first term, and \(b=2 y\) is the second term. For the 4 th term, \(r+1=4\), which means \(r=3\).
Step 2: Apply the general term formula
Substitute these values into the general term formula:
\(
\mathrm{T}_4={ }^5 \mathrm{C}_3(3 x)^{5-3}(2 y)^3
\)
Step 3: Calculate the combination coefficient
Calculate the binomial coefficient \({ }^n \mathrm{C}_{\boldsymbol{r}}\) :
\(
{ }^5 C_3=\frac{5!}{3!(5-3)!}=\frac{5!}{3!2!}=10
\)
Step 4: Simplify the powers and combine terms
Simplify the terms with exponents and multiply all coefficients:
\(
T_4=10 \cdot(3 x)^2 \cdot(2 y)^3=10 \cdot\left(9 x^2\right) \cdot\left(8 y^3\right)=720 x^2 y^3
\)

Middle term in the expansion \((a+b)^n\), we have

Case I: If \(n\) is even, then the number of terms in the expansion will be \(n+1\). Since \(n\) is even so \(n+1\) is odd. Therefore, the middle term is \(\left(\frac{n+1+1}{2}\right)^{t h}\), i.e., \(\left(\frac{n}{2}+1\right)^{t h}\) term.

Reasoning: Consider the binomial expansion of \((a+b)^n\).
It is known that the total number of terms in the expansion of \((a+b)^n\) is \(n+1\).
If \(n\) is even, then \(n+1\) is odd. Hence, the expansion contains an odd number of terms.
In any sequence with an odd number of terms, there exists exactly one middle term. The position of the middle term is
\(
\frac{\text { number of terms }+1}{2}=\frac{(n+1)+1}{2}=\frac{n+2}{2}=\left(\frac{n}{2}+1\right) .
\)
Therefore, when \(n\) is even, the middle term of the expansion of \((a+b)^n\) is the
\(
\left(\frac{n}{2}+1\right)^{\text {th }} \text { term. }
\)

For example: \(n=6\) (even)
\(
(a+b)^6=a^6+6 a^5 b+15 a^4 b^2+20 a^3 b^3+15 a^2 b^4+6 a b^5+b^6
\)
Number of terms \(=6+1=7\) (odd)
Middle term position:
\(
\frac{6}{2}+1=4
\)
So the 4th term is the middle term:
\(
20 a^3 b^3
\)

For example, in the expansion of \((x+2 y)^8\), the middle term is \(\left(\frac{8}{2}+1\right)^{t h}\) i.e., \(5^{\text {th }}\) term.
These examples confirm that when \(n\) is even, the middle term is always the \(\left(\frac{n}{2}+1\right)^{\text {th }}\) term.

Case-II: If \(n\) is odd, then \(n+1\) is even, so there will be two middle terms in the expansion, namely, \(\left(\frac{n+1}{2}\right)^{t h}\) term and \(\left(\frac{n+1}{2}+1\right)^{t h}=\left(\frac{n+3}{2}\right)^{t h}\) term.

Reasoning: Consider the binomial expansion of \((a+b)^n\).
The total number of terms in the expansion of \((a+b)^n\) is \(n+1\).
If \(n\) is odd, then \(n+1\) is even. Hence, the expansion contains an even number of terms.
In any sequence with an even number of terms, there are two middle terms. If the total number of terms is \(n+1\), then the positions of the two middle terms are
\(
\frac{n+1}{2} \text { and }\left(\frac{n+1}{2}+1\right) .
\)
Therefore, when \(n\) is odd, the two middle terms in the expansion of \((a+b)^n\) are the
\(
\left(\frac{n+1}{2}\right)^{\text {th }} \text { term and }\left(\frac{n+1}{2}+1\right)^{\text {th }} \text { term. }
\)
Hence proved.

For example:
Let \(n=5\) (odd).
\(
(a+b)^5=a^5+5 a^4 b+10 a^3 b^2+10 a^2 b^3+5 a b^4+b^5
\)
Number of terms \(=5+1=6\) (even)
Middle terms:
\(
\frac{6}{2}=3 \text { and } 4
\)
So the 3rd and 4th terms are the middle terms:
\(
10 a^3 b^2 \text { and } 10 a^2 b^3
\)
This confirms the result.

Similarly, in the expansion \((2 x-y)^7\), the middle terms are \(\left(\frac{7+1}{2}\right)^{\text {th }}\), i.e., \(4^{\text {th }}\) and \(\left(\frac{7+1}{2}+1\right)^{t h}\), i.e., \(5^{\text {th }}\) term.

Greatest binomial coefficient

Case-I (when \(n\) is an even): Middle term always carries the greatest binomial coefficient. As when \(n\) is an even middle term, \(T_{n / 2+1}\) has the greatest binomial coefficient \({ }^n C_{n / 2}\).

Proof: Consider the binomial expansion of \((a+b)^n\).
The general term is
\(
T_{r+1}={ }^n C_r a^{n-r} b^r, \quad r=0,1,2, \ldots, n.
\)
So the binomial coefficients appearing are
\(
{ }^n C_0,{ }^n C_1,{ }^n C_2, \ldots,{ }^n C_n.
\)
Step 1: Behaviour of consecutive binomial coefficients
Consider the ratio of two successive binomial coefficients:
\(
\frac{{ }^n C_{r+1}}{{ }^n C_r}=\frac{n-r}{r+1}.
\)
If \(r<\frac{n}{2}\), then \(n-r>r+1\), so
\(
\frac{{ }^n C_{r+1}}{{ }^n C_r}>1,
\)
which means \({ }^n C_{r+1}>{ }^n C_r\).
If \(r=\frac{n}{2}\) (with \(n\) even), then
\(
\frac{{ }^n C_{r+1}}{{ }^n C_r}=1,
\)
so the coefficients stop increasing at this point.
If \(r>\frac{n}{2}\), then \(n-r<r+1\), so
\(
\frac{{ }^n C_{r+1}}{{ }^n C_r}<1,
\)
and the coefficients start decreasing.
Step 2: Greatest binomial coefficient
Hence, the binomial coefficients increase as \(r\) goes from 0 to \(\frac{n}{2}\) and decrease thereafter.
Therefore, when \(n\) is even, the greatest binomial coefficient is
\(
{ }^n C_{n / 2}.
\)
Step 3: Middle term
For even \(n\), the middle term of the expansion is
\(
T_{\frac{n}{2}+1}.
\)
The binomial coefficient in this term is \({ }^n C_{n / 2}\).
Thus, the middle term always carries the greatest binomial coefficient, and when \(n\) is even, the middle term \(T_{\frac{n}{2}+1}\) has the greatest binomial coefficient
\(
{ }^n C_{n / 2}.
\)

For example: \(n=6\) (even)
Binomial expansion:
\(
(a+b)^6=a^6+6 a^5 b+15 a^4 b^2+20 a^3 b^3+15 a^2 b^4+6 a b^5+b^6
\)
Binomial coefficients:
\(
1,6,15,20,15,6,1
\)
Middle term position:
\(
\frac{6}{2}+1=4
\)
Middle term: \(20 a^3 b^3\)
Greatest binomial coefficient: \({ }^6 C_3=20\).

Case-II (when \(n\) is an odd): when \(n\) is an odd middle term, \(T_{(n+1) / 2}\) and \(T_{(n+3) / 2}\) or \(T_{\left(\frac{n-1}{2}\right)+1}\) and \(T_{\left(\frac{n+1}{2}\right)+1}\) have the greatest binomial coefficients \({ }^n C_{\left(\frac{n-1}{2}\right)}\) and \({ }^n C_{\left(\frac{n+1}{2}\right)}\).

Proof: Consider the binomial expansion of \((a+b)^n\), where \(n\) is odd.
The general term is
\(
T_{r+1}={ }^n C_r a^{n-r} b^r, \quad r=0,1,2, \ldots, n.
\)
Thus, the binomial coefficients appearing are
\(
{ }^n C_0,{ }^n C_1,{ }^n C_2, \ldots,{ }^n C_n.
\)
Step 1: Compare successive binomial coefficients
Consider the ratio:
\(
\frac{{ }^n C_{r+1}}{{ }^n C_r}=\frac{n-r}{r+1}.
\)
If \(r<\frac{n-1}{2}\), then \(n-r>r+1\), so
\(
{ }^n C_{r+1}>{ }^n C_r.
\)
If \(r=\frac{n-1}{2}\), then
\(
\frac{{ }^n C_{r+1}}{{ }^n C_r}=\frac{n-\frac{n-1}{2}}{\frac{n-1}{2}+1}=1,
\)
hence
\(
{ }^n C_{\frac{n-1}{2}}={ }^n C_{\frac{n+1}{2}} .
\)
If \(r>\frac{n-1}{2}\), then \(n-r<r+1\), so
\(
{ }^n C_{r+1}<{ }^n C_r.
\)
Thus, the binomial coefficients increase, become maximum at two consecutive terms, and then decrease.
Step 2: Identify the middle terms
When \(n\) is odd, the total number of terms is \(n+1\), which is even.
Hence, there are two middle terms:
\(
T_{\frac{n+1}{2}} \text { and } T_{\frac{n+3}{2}} .
\)
These correspond to:
\(
r=\frac{n-1}{2} \quad \text { and } \quad r=\frac{n+1}{2} .
\)
Step 3: Greatest binomial coefficients
The binomial coefficients in these two middle terms are:
\(
{ }^n C_{\frac{n-1}{2}} \text { and }{ }^n C_{\frac{n+1}{2}},
\)
and from Step 1, these two are equal and greater than all other binomial coefficients.
Therefore, when \(n\) is odd, the two middle terms
\(
T_{\frac{n+1}{2}} \text { and } T_{\frac{n+3}{2}}
\)
(or equivalently \(T_{\left(\frac{n-1}{2}\right)+1}\) and \(T_{\left(\frac{n+1}{2}\right)+1}\) ) carry the greatest binomial coefficients
\(
{ }^n C_{\frac{n-1}{2}} \text { and }{ }^n C_{\frac{n+1}{2}} \text {. }
\)

For example: \(n=7\) (odd)
\(
(a+b)^7=a^7+7 a^6 b+21 a^5 b^2+35 a^4 b^3+35 a^3 b^4+21 a^2 b^5+7 a b^6+b^7
\)
Binomial coefficients:
\(
1,7,21,35,35,21,7,1
\)
Middle terms:
\(
T_4 \text { and } T_5
\)
Greatest binomial coefficients:
\(
{ }^7 C_3={ }^7 C_4=35
\)

Example 2: In the expansion of \(\left(x+\frac{1}{x}\right)^{2 n}\), where \(x \neq 0\), show that the middle term is \(\left(\frac{2 n+1+1}{2}\right)^{t h}\), i.e., \((n+1)^{\text {th }}\) term, as \(2 n\) is even. It is given by \({ }^{2 n} \mathrm{C}_n x^n\left(\frac{1}{x}\right)^n={ }^{2 n} \mathrm{C}_n\) (constant). This term is called the term independent of \(x\) or the constant term.

Solution: Consider the binomial expansion of
\(
\left(x+\frac{1}{x}\right)^{2 n}, \quad x \neq 0.
\)
Step 1: Number and position of the middle term
Since the exponent is \(2 n\), the total number of terms in the expansion is
\(
2 n+1 .
\)
Because \(2 n\) is even, \(2 n+1\) is odd, so there is one middle term.
The position of the middle term is
\(
\frac{(2 n+1)+1}{2}=\frac{2 n+2}{2}=n+1.
\)
Hence, the middle term is the \((n+1)^{\text {th }}\) term.
Step 2: General term of the expansion
The general term of \(\left(x+\frac{1}{x}\right)^{2 n}\) is
\(
T_{r+1}={ }^{2 n} C_r x^{2 n-r}\left(\frac{1}{x}\right)^r,
\)
where \(r=0,1,2, \ldots, 2 n\).
Simplifying,
\(
T_{r+1}={ }^{2 n} C_r x^{2 n-2 r} .
\)
Step 3: Middle term
The middle term corresponds to
\(
r+1=n+1 \quad \Rightarrow \quad r=n .
\)
Substitute \(r=n\) into the general term:
\(
T_{n+1}={ }^{2 n} C_n x^{2 n-2 n}={ }^{2 n} C_n x^0={ }^{2 n} C_n
\)

Example 3: Find \(a\) if the \(17^{\text {th }}\) and \(18^{\text {th }}\) terms of the expansion \((2+a)^{50}\) are equal.

Solution: The \((r+1)^{t h}\) term of the expansion \((x+y)^n\) is given by \(\mathrm{T}_{r+1}={ }^n \mathrm{C}_r x^{n-r} y^r\).
For the \(17^{\text {th }}\) term, we have, \(r+1=17\), i.e., \(r=16\)
Therefore, \(\quad \mathrm{T}_{17}=\mathrm{T}_{16+1}={ }^{50} \mathrm{C}_{16}(2)^{50-16} a^{16}\)
\(
={ }^{50} \mathrm{C}_{16} 2^{34} a^{16} \text {. }
\)
Similarly, \(\quad \mathrm{T}_{18}={ }^{50} \mathrm{C}_{17} 2^{33} a^{17}\)
Given that \(\quad \mathrm{T}_{17}=\mathrm{T}_{18}\)
So \({ }^{50} \mathrm{C}_{16}(2)^{34} a^{16}={ }^{50} \mathrm{C}_{17}(2)^{33} a^{17}\)
Therefore
\(\frac{{ }^{50} \mathrm{C}_{16} \cdot 2^{34}}{{ }^{50} \mathrm{C}_{17} \cdot 2^{33}}=\frac{a^{17}}{a^{16}}\)
i.e., \(\quad a=\frac{{ }^{50} \mathrm{C}_{16} \times 2}{{ }^{50} \mathrm{C}_{17}}=\frac{50 !}{16 ! 34 !} \times \frac{17 ! \cdot 33 !}{50 !} \times 2=1\)

Example 4: Show that the middle term in the expansion of \((1+x)^{2 n}\) is \(\frac{1.3 .5 \ldots(2 n-1)}{n !} 2^n x^n\), where \(n\) is a positive integer.

Solution: As \(2 n\) is even, the middle term of the expansion \((1+x)^{2 n}\) is \(\left(\frac{2 n}{2}+1\right)^{\text {th }}\), i.e., \((n+1)^{\mathrm{th}}\) term which is given by,
\(
\begin{aligned}
\mathrm{T}_{n+1} & ={ }^{2 n} \mathrm{C}_n(1)^{2 n-n}(x)^n={ }^{2 n} \mathrm{C}_n x^n=\frac{(2 n) !}{n ! n !} x^n \\
& =\frac{2 n(2 n-1)(2 n-2) \ldots 4.3 .2 .1}{n ! n !} x^n \\
& =\frac{1.2 .3 .4 \ldots(2 n-2)(2 n-1)(2 n)}{n ! n !} x^n \\
& =\frac{[1.3 .5 \ldots(2 n-1)][2.4 .6 \ldots(2 n)]}{n ! n !} x^n \\
& =\frac{[1.3 .5 \ldots(2 n-1)] 2^n[1.2 .3 \ldots n]}{n ! n !} x^n \\
& =\frac{[1.3 .5 \ldots(2 n-1)] n !}{n ! n !} 2^n \cdot x^n \\
& =\frac{1.3 .5 \ldots(2 n-1)}{n !} 2^n x^n
\end{aligned}
\)

Example 5: Find the coefficient of \(x^6 y^3\) in the expansion of \((x+2 y)^9\).

Solution: Suppose \(x^6 y^3\) occurs in the \((r+1)^{\text {th }}\) term of the expansion \((x+2 y)^9\). Now \(\quad \mathrm{T}_{r+1}={ }^9 \mathrm{C}_r x^{9-r}(2 y)^r={ }^9 \mathrm{C}_r 2^r \cdot x^{9-r} \cdot y^r\).
Comparing the indices of \(x\) as well as \(y\) in \(x^6 y^3\) and in \(\mathrm{T}_{r+1}\), we get \(r=3\). Thus, the coefficient of \(x^6 y^3\) is
\(
{ }^9 \mathrm{C}_3 2^3=\frac{9 !}{3 ! 6 !} \cdot 2^3=\frac{9 \cdot 8 \cdot 7}{3 \cdot 2} \cdot 2^3=672
\)

Example 6: The second, third and fourth terms in the binomial expansion \((x+a)^n\) are 240,720 and 1080, respectively. Find \(x, a\) and \(n\).

Solution: Given that second term \(\mathrm{T}_2=240\)
We have \(\quad \mathrm{T}_2={ }^n \mathrm{C}_1 x^{n-1} \cdot a\)
So \(\quad{ }^n \mathrm{C}_1 x^{n-1} \cdot a=240 \dots(1)\)
Similarly
\({ }^n \mathrm{C}_2 x^{n-2} a^2=720 \dots(2)\)
and
\({ }^n \mathrm{C}_3 x^{n-3} a^3=1080 \dots(3)\)
Dividing (2) by (1), we get
or \(\frac{{ }^n \mathrm{C}_2 x^{n-2} a^2}{{ }^n \mathrm{C}_1 x^{n-1} a}=\frac{720}{240}\) i.e., \(\quad \frac{(n-1) !}{(n-2) !} \cdot \frac{a}{x}=6\); \(\frac{a}{x}=\frac{6}{(n-1)} \dots(4)\)
Dividing (3) by (2), we have
\(
\frac{a}{x}=\frac{9}{2(n-2)} \dots(5)
\)
From (4) and (5),
\(
\frac{6}{n-1}=\frac{9}{2(n-2)} . \quad \text { Thus, } n=5
\)
Hence, from (1), \(5 x^4 a=240\), and from (4), \(\frac{a}{x}=\frac{3}{2}\)
Solving these equations for \(a\) and \(x\), we get \(x=2\) and \(a=3\).

Example 7: The coefficients of three consecutive terms in the expansion of \((1+a)^n\) are in the ratio1: \(7: 42\). Find \(n\).

Solution: Suppose the three consecutive terms in the expansion of \((1+a)^n\) are \((r-1)^{\mathrm{th}}, r^{\mathrm{th}}\) and \((r+1)^{\mathrm{th}}\) terms.

The \((r-1)^{\mathrm{th}}\) term is \({ }^n \mathrm{C}_{r-2} a^{r-2}\), and its coefficient is \({ }^n \mathrm{C}_{r-2}\).
Similarly, the coefficients of \(r^{\text {th }}\) and \((r+1)^{\text {th }}\) terms are \({ }^n \mathrm{C}_{r-1}\) and \({ }^n \mathrm{C}_r\), respectively.
Since the coefficients are in the ratio \(1: 7: 42\), so we have,
\(\frac{{ }^n \mathrm{C}_{r-2}}{{ }^n \mathrm{C}_{r-1}}=\frac{1}{7}\), i.e., \(n-8 r+9=0 \dots(1)\)
and
\(\frac{{ }^n \mathrm{C}_{r-1}}{{ }^n \mathrm{C}_r}=\frac{7}{42}\), i.e., \(n-7 r+1=0 \dots(2)\)
Solving equations(1) and (2), we get, \(n=55\).

Example 8: \(\text { Find the term independent of } x \text { in the expansion of }\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^6 \text {. }
\)

Solution: We have \(\mathrm{T}_{r+1}={ }^6 \mathrm{C}_r\left(\frac{3}{2} x^2\right)^{6-r}\left(-\frac{1}{3 x}\right)^r\)
\(
={ }^6 \mathrm{C}_r\left(\frac{3}{2}\right)^{6-r}\left(x^2\right)^{6-r}(-1)^r\left(\frac{1}{x}\right)^r\left(\frac{1}{3^r}\right)
\)
\(
=(-1)^{r \quad 6} \mathrm{C}_r \frac{(3)^{6-2 r}}{(2)^{6-r}} x^{12-3 r}
\)
The term will be independent of \(x\) if the index of \(x\) is zero, i.e., \(12-3 r=0\). Thus, \(r=4\) Hence \(5^{\text {th }}\) term is independent of \(x\) and is given by \((-1)^4{ }^6 \mathrm{C}_4 \frac{(3)^{6-8}}{(2)^{6-4}}=\frac{5}{12}\).

Example 9: If the coefficients of \(a^{r-1}, a^r\) and \(a^{r+1}\) in the expansion of \((1+a)^n\) are in arithmetic progression, prove that \(n^2-n(4 r+1)+4 r^2-2=0\).

Solution: The \((r+1)^{\mathrm{th}}\) term in the expansion is \({ }^n \mathrm{C}_r a^r\). Thus it can be seen that \(a^r\) occurs in the \((r+1)^{\text {th }}\) term, and its coefficient is \({ }^n \mathrm{C}_r\). Hence the coefficients of \(a^{r-1}, a^r\) and \(a^{r+1}\) are \({ }^n \mathrm{C}_{r-1},{ }^n \mathrm{C}_r\) and \({ }^n \mathrm{C}_{r+1}\), respectively. Since these coefficients are in arithmetic progression, so we have, \({ }^n \mathrm{C}_{r-1}+{ }^n \mathrm{C}_{r+1}=2 .{ }^n \mathrm{C}_r\). This gives
i.e.
\(
\begin{aligned}
& \frac{n !}{(r-1) !(n-r+1) !}+\frac{n !}{(r+1) !(n-r-1) !}=2 \times \frac{n !}{r !(n-r) !} \\
& \frac{1}{(r-1) !(n-r+1)(n-r)(n-r-1) !}+\frac{1}{(r+1)(r)(r-1) !(n-r-1) !} \\
& =2 \times \frac{1}{r(r-1) !(n-r)(n-r-1) !} \\
& \frac{1}{(r-1) !(n-r-1) !}\left[\frac{1}{(n-r)(n-r+1)}+\frac{1}{(r+1)(r)}\right] \\
& =2 \times \frac{1}{(r-1) !(n-r-1) ![r(n-r)]} \\
& \frac{1}{(n-r+1)(n-r)}+\frac{1}{r(r+1)}=\frac{2}{r(n-r)}, \\
& \frac{r(r+1)+(n-r)(n-r+1)}{(n-r)(n-r+1) r(r+1)}=\frac{2}{r(n-r)} \\
& r(r+1)+(n-r)(n-r+1)=2(r+1)(n-r+1) \\
& r^2+r+n^2-n r+n-n r+r^2-r=2\left(n r-r^2+r+n-r+1\right)
\end{aligned}
\)

or \(\quad n^2-4 n r-n+4 r^2-2=0\)
i.e., \(\quad n^2-n(4 r+1)+4 r^2-2=0\)

Example 10: Show that the coefficient of the middle term in the expansion of \((1+x)^{2 n}\) is equal to the sum of the coefficients of two middle terms in the expansion of \((1+x)^{2 n-1}\). 

Solution: As \(2 n\) is even so the expansion \((1+x)^{2 n}\) has only one middle term which is \(\left(\frac{2 n}{2}+1\right)^{\text {th }}\) i.e., \((n+1)^{\text {th }}\) term.
The \((n+1)^{\text {th }}\) term is \({ }^{2 n} \mathrm{C}_n x^n\). The coefficient of \(x^n\) is \({ }^{2 n} \mathrm{C}_n\) Similarly, \((2 n-1)\) being odd, the other expansion has two middle terms, \(\left(\frac{2 n-1+1}{2}\right)^{\text {th }}\) and \(\left(\frac{2 n-1+1}{2}+1\right)^{\text {th }}\) i.e., \(n^{\text {th }}\) and \((n+1)^{\text {th }}\) terms. The coefficients of these terms are \({ }^{2 n-1} \mathrm{C}_{n-1}\) and \({ }^{2 n-1} \mathrm{C}_n\), respectively.
Now
\(
{ }^{2 n-1} \mathrm{C}_{n-1}+{ }^{2 n-1} \mathrm{C}_n={ }^{2 n} \mathrm{C}_n \quad\left[\mathrm{As}{ }^n \mathrm{C}_{r-1}+{ }^n \mathrm{C}_r={ }^{n+1} \mathrm{C}_r\right] \text {. as required. }
\)

Example 11: Find the coefficient of \(a^4\) in the product \((1+2 a)^4(2-a)^5\) using binomial theorem.

Solution: We first expand each of the factors of the given product using Binomial Theorem. We have
\(
\begin{aligned}
(1+2 a)^4= & { }^4 \mathrm{C}_0+{ }^4 \mathrm{C}_1(2 a)+{ }^4 \mathrm{C}_2(2 a)^2+{ }^4 \mathrm{C}_3(2 a)^3+{ }^4 \mathrm{C}_4(2 a)^4 \\
= & 1+4(2 a)+6\left(4 a^2\right)+4\left(8 a^3\right)+16 a^4 . \\
= & 1+8 a+24 a^2+32 a^3+16 a^4 \\
\text { and }(2-a)^5= & { }^5 \mathrm{C}_0(2)^5-{ }^5 \mathrm{C}_1(2)^4(a)+{ }^5 \mathrm{C}_2(2)^3(a)^2-{ }^5 \mathrm{C}_3(2)^2(a)^3 \\
& +{ }^5 \mathrm{C}_4(2)(a)^4-{ }^5 \mathrm{C}_5(a)^5 \\
= & 32-80 a+80 a^2-40 a^3+10 a^4-a^5
\end{aligned}
\)
Thus \((1+2 a)^4(2-a)^5\)
\(
=\left(1+8 a+24 a^2+32 a^3+16 a^4\right)\left(32-80 a+80 a^2-40 a^3+10 a^4-a^5\right)
\)
The complete multiplication of the two brackets need not be carried out. We write only those terms which involve \(a^4\). This can be done if we note that \(a^r . a^{4-r}=a^4\). The terms containing \(a^4\) are
\(
1\left(10 a^4\right)+(8 a)\left(-40 a^3\right)+\left(24 a^2\right)\left(80 a^2\right)+\left(32 a^3\right)(-80 a)+\left(16 a^4\right)(32)=-438 a^4
\)
\(
\text { Thus, the coefficient of } a^4 \text { in the given product is }-438 \text {. }
\)

Example 12: Find the \(r^{\text {th }}\) term from the end in the expansion of \((x+a)^n\).

Solution: There are \((n+1)\) terms in the expansion of \((x+a)^n\). Observing the terms we can say that the first term from the end is the last term, i.e., \((n+1)^{\text {th }}\) term of the expansion and \(n+1=(n+1)-(1-1)\).
The second term from the end is the \(n^{\text {th }}\) term of the expansion, and \(n=(n+1)-(2-1)\).
The third term from the end is the \((n-1)^{\text {th }}\) term of the expansion and \(n-1=(n+1)-(3-1)\) and so on.
Thus \(r^{\text {th }}\) term from the end will be term number \((n+1)-(r-1)=(n-r+2)\) of the expansion.
And the \((n-r+2)^{\text {th }}\) term is \({ }^n \mathrm{C}_{n-r+1} x^{r-1} a^{n-r+1}\).

Example 13: \(\text { Find the term independent of } x \text { in the expansion of }\left(\sqrt[3]{x}+\frac{1}{2 \sqrt[3]{x}}\right)^{18}, x>0 \text {. }\)

Solution: We have \(\mathrm{T}_{r+1}={ }^{18} \mathrm{C}_r(\sqrt[3]{x})^{18-r}\left(\frac{1}{2 \sqrt[3]{x}}\right)^r\)
\(
={ }^{18} \mathrm{C}_r x^{\frac{18-r}{3}} \cdot \frac{1}{2^r \cdot x^{\frac{r}{3}}}={ }^{18} \mathrm{C}_r \frac{1}{2^r} \cdot x^{\frac{18-2 r}{3}}
\)
Since we have to find a term independent of \(x\), i.e., term not having \(x\), so take \(\frac{18-2 r}{3}=0\).
We get \(r=9\). The required term is \({ }^{18} \mathrm{C}_9 \frac{1}{2^9}\).

Example 14: The sum of the coefficients of the first three terms in the expansion of \(\left(x-\frac{3}{x^2}\right)^m, x \neq 0, m\) being a natural number, is 559. Find the term of the expansion containing \(x^3\).

Solution: The coefficients of the first three terms of \(\left(x-\frac{3}{x^2}\right)^m\) are \({ }^m \mathrm{C}_0,(-3)^m \mathrm{C}_1\) and \(9{ }^m C_2\). Therefore, by the given condition, we have
\(
{ }^m \mathrm{C}_0-3^m \mathrm{C}_1+9^m \mathrm{C}_2=559 \text {, i.e., } 1-3 m+\frac{9 m(m-1)}{2}=559
\)
which gives \(m=12\) ( \(m\) being a natural number).
Now \(\quad \mathrm{T}_{r+1}={ }^{12} \mathrm{C}_r x^{12-r}\left(-\frac{3}{x^2}\right)^r={ }^{12} \mathrm{C}_r(-3)^r \cdot x^{12-3 r}\)
Since we need the term containing \(x^3\), so put \(12-3 r=3\) i.e., \(r=3\).
Thus, the required term is \({ }^{12} \mathrm{C}_3(-3)^3 x^3\), i.e., \(-5940 x^3\).

Example 15: If the coefficients of \((r-5)^{\mathrm{th}}\) and \((2 r-1)^{\mathrm{th}}\) terms in the expansion of \((1+x)^{34}\) are equal, find \(r\).

Solution: The coefficients of \((r-5)^{\mathrm{th}}\) and \((2 r-1)^{\text {th }}\) terms of the expansion \((1+x)^{34}\) are \({ }^{34} \mathrm{C}_{r-6}\) and \({ }^{34} \mathrm{C}_{2 r-2}\), respectively.
Since they are equal so \({ }^{34} \mathrm{C}_{r-6}={ }^{34} \mathrm{C}_{2 r-2}\)
Therefore, either \(r-6=2 r-2\) or \(r-6=34-(2 r-2)\)
[Using the fact that if \({ }^n \mathrm{C}_r={ }^n \mathrm{C}_p\), then either \(r=p\) or \(r=n-p\) ]
So, we get \(r=-4\) or \(r=14\). \(r\) being a natural number, \(r=-4\) is not possible.
So, \(r=14\).

Example 16: If the middle term in the expansion of \(\left(x^2+1 / x\right)^n\) is \(924 x^6\), then find the value of \(n\).

Solution: Since there is only one middle term here, \(n\) is even, therefore \((\boldsymbol{n} / 2+1)^{\text {th }}\) term is the middle term. Hence,
\(
\begin{aligned}
& { }^n C_{n / 2}\left(x^2\right)^{n / 2}\left(\frac{1}{x}\right)^{n / 2}=924 x^6 \\
\Rightarrow & x^{n / 2}=x^6 \\
\Rightarrow & n=12\left(\text { also }{ }^{12} C_6=924\right)
\end{aligned}
\)

Explanation: Step 1: Nature of the middle term
Since only one middle term is mentioned, the exponent \(n\) must be even.
Hence, the middle term is the
\(
\left(\frac{n}{2}+1\right)^{\text {th }} \text { term. }
\)
Step 2: General term of the expansion
The general term of
\(
\left(x^2+\frac{1}{x}\right)^n
\)
is \(T_{r+1}={ }^n C_r\left(x^2\right)^{n-r}\left(\frac{1}{x}\right)^r\)
Step 3: Middle term
For the middle term,
\(
r=\frac{n}{2}
\)
Substitute \(r=\frac{n}{2}\) :
\(
T_{\frac{n}{2}+1}={ }^n C_{n / 2}\left(x^2\right)^{n / 2}\left(\frac{1}{x}\right)^{n / 2} .
\)
Simplify:
\(
\left(x^2\right)^{n / 2}=x^n, \quad\left(\frac{1}{x}\right)^{n / 2}=x^{-n / 2}
\)
So, \(T_{\frac{n}{2}+1}={ }^n C_{n / 2} x^{n-n / 2}={ }^n C_{n / 2} x^{n / 2}\)
Step 4: Compare with the given middle term
Given middle term:
\(
924 x^6
\)
Equating powers of \(x\) :
\(
\frac{n}{2}=6 \Rightarrow n=12
\)
Step 5: Verify the coefficient
\(
{ }^{12} C_6=924,
\)
which matches the given coefficient.

Example 17: If the coefficient of the middle term in the expansion of \((1+x)^{2 n+2}\) is \(\alpha\) and the coefficients of middle terms in the expansion of \((1+x)^{2 n+1}\) are \(\beta\) and \(\gamma\), then relate \(\alpha, \beta\) and \(\gamma\).

Solution: Since \((n+2)^{\text {th }}\) term is the middle term in the expansion of \((1+x)^{2 n+2}\), therefore \(\alpha={ }^{2 n+2} C_{n+1}\). Since \((n+1)^{\text {th }}\) and \((n+2)^{\text {th }}\) terms are middle terms in the expansion of \((1+x)^{2 n+1}\), therefore,
\(
\beta={ }^{2 n+1} C_n \text { and } \gamma={ }^{2 n+1} C_{n+1}
\)
But
\(
{ }^{2 n+1} C_n+{ }^{2 n+1} C_{n+1}={ }^{2 n+2} C_{n+1} \Rightarrow \beta+\gamma=\alpha
\)

Explanation: We are asked to relate the coefficients of the middle terms in binomial expansions:
\((1+x)^{2 n+2} \rightarrow\) middle term coefficient is \(\alpha\)
\((1+x)^{2 n+1} \rightarrow\) middle term coefficients are \(\beta\) and \(\gamma\)
Step 1: Middle term of \((1+x)^{2 n+2}\)
Here, \(2 n+2\) is even, so there is a single middle term.
Number of terms \(=2 n+3 \rightarrow\) odd, so middle term position:
\(
\frac{(2 n+3)+1}{2}=\frac{2 n+4}{2}=n+2
\)
Coefficient of middle term:
\(
\alpha={ }^{2 n+2} C_{n+1}
\)
Step 2: Middle terms of \((1+x)^{2 n+1}\)
Here, \(2 n+1\) is odd, so there are two middle terms.
Number of terms \(=2 n+2 \rightarrow\) even, so the two middle terms are at positions:
\(
T_{\frac{2 n+2}{2}}=T_{n+1} \quad \text { and } \quad T_{\frac{2 n+2}{2}+1}=T_{n+2}
\)
Coefficients:
\(
\beta={ }^{2 n+1} C_n, \quad \gamma={ }^{2 n+1} C_{n+1}
\)
Step 3: Relating \(\alpha, \beta, \gamma\)
Use the Pascal’s identity:
\(
{ }^{m+1} C_r={ }^m C_{r-1}+{ }^m C_r
\)
Apply it with \(m=2 n+1\) and \(r=n+1\) :
\(
{ }^{2 n+2} C_{n+1}={ }^{2 n+1} C_n+{ }^{2 n+1} C_{n+1}
\)
But from Step 1-2:
\(
\alpha={ }^{2 n+2} C_{n+1}, \quad \beta={ }^{2 n+1} C_n, \quad \gamma={ }^{2 n+1} C_{n+1}
\)
Hence:
\(
\alpha=\beta+\gamma
\)
The relationship between the coefficients is:
\(
\alpha=\beta+\gamma
\)
Remark: This shows that the middle term coefficient of the expansion with even exponent equals the sum of the two middle term coefficients of the expansion with odd exponent.

Ratio of Consecutive Terms/Coefficients

Coefficients of \(x^r\) and \(x^{r+1}\) are \({ }^h C_{r-1}\) and \({ }^h C_r\), respectively. Also, we know that
\(
\frac{{ }^n C_r}{{ }^n C_{r-1}}=\frac{n-r+1}{r}
\)
Similarly,
\(
\frac{{ }^n C_{r+1}}{{ }^n C_r}=\frac{n-r}{r+1} \text { (replacing } r \text { by } r+1)
\)
\(\frac{{ }^{n+1} C_{r+1}}{{ }^{n+1} C_r}=\frac{n-r+1}{r+1}(\) replacing \(r\) by \(r+1\) and \(n\) by \(n+1)\) and so on.

Derivation: Ratio of consecutive coefficients in \((1+x)^n\)
The general term of \((1+x)^n\) is:
\(
T_{r+1}={ }^n C_r x^r, \quad r=0,1,2, \ldots, n
\)
The ratio of consecutive coefficients is:
\(
\frac{{ }^n C_r}{{ }^n C_{r-1}}=\frac{n-r+1}{r}, \quad r=1,2, \ldots, n
\)
Derivation:
\(
{ }^n C_r=\frac{n!}{r!(n-r)!}, \quad{ }^n C_{r-1}=\frac{n!}{(r-1)!(n-r+1)!}
\)
So,
\(
\frac{{ }^n C_r}{{ }^n C_{r-1}}=\frac{n!/(r!(n-r)!)}{n!/((r-1)!(n-r+1)!)}=\frac{n-r+1}{r} \dots(1)
\)
This tells us how the coefficients increase or decrease as \(r\) increases.

Ratio of next-to-next coefficients:
Similarly, the ratio of the \((r+1)^{\text {th }}\) coefficient to the \(r^{\text {th }}\) coefficient is:
\(
\frac{{ }^n C_{r+1}}{{ }^n C_r}=\frac{n-r}{r+1}, \quad r=0,1,2, \ldots, n-1 \dots(2)
\)
This comes from replacing \(r\) by \(r+1\) in equation 1.

Ratio for \((1+x)^{n+1}\):
For \((1+x)^{n+1}\), the same formula applies:
\(
\frac{{ }^{n+1} C_{r+1}}{{ }^{n+1} C_r}=\frac{(n+1)-r}{r+1}=\frac{n-r+1}{r+1}
\)
This is obtained by replacing \(n\) by \(n+1\) in equation 2.

Example 18: If the coefficients of three consecutive terms in the expansion of \((1+x)^n\) are 165, 330 and 462, respectively, then find the value of \(n\).

Solution: We are given three consecutive coefficients in \((1+x)^n\) :
\(
{ }^n C_{r-1}=165, \quad{ }^n C_r=330, \quad{ }^n C_{r+1}=462
\)
We are asked to find \(n\).
Step 1: Use ratio of consecutive coefficients
We know:
\(
\frac{{ }^n C_r}{{ }^n C_{r-1}}=\frac{n-r+1}{r}
\)
Substitute the given numbers:
\(
\frac{330}{165}=2=\frac{n-r+1}{r}
\)
Solve for \(n\) in terms of \(r\) :
\(
2=\frac{n-r+1}{r} \quad \Rightarrow \quad n-r+1=2 r \quad \Rightarrow \quad n=3 r-1
\)
Step 2: Use ratio for the next pair
Similarly:
\(
\frac{{ }^n C_{r+1}}{{ }^n C_r}=\frac{n-r}{r+1}
\)
Substitute the given numbers:
\(
\frac{462}{330}=\frac{21}{15}=\frac{7}{5}=\frac{n-r}{r+1}
\)
Step 3: Solve for \(n\) using this ratio
\(
\begin{aligned}
\frac{n-r}{r+1}=\frac{7}{5} & \Rightarrow 5(n-r)=7(r+1) \\
5 n-5 r=7 r+7 & \Rightarrow 5 n=12 r+7 \quad \Rightarrow \quad n=\frac{12 r+7}{5}
\end{aligned}
\)
Step 4: Equate the two expressions for \(n\)
From Step 1: \(n=3 r-1\)
From Step 3: \(n=\frac{12 r+7}{5}\)
\(
3 r-1=\frac{12 r+7}{5}
\)
Multiply both sides by 5 :
\(
\begin{gathered}
15 r-5=12 r+7 \\
15 r-12 r=7+5 \Rightarrow 3 r=12 \quad \Rightarrow \quad r=4
\end{gathered}
\)
Step 5: Find \(n\)
\(
n=3 r-1=3(4)-1=12-1=11
\)
Check: The coefficients of consecutive terms for \(n=11\) and \(r=4\) are:
\(
{ }^{11} C_3=165, \quad{ }^{11} C_4=330, \quad{ }^{11} C_5=462
\)

Example 19: Find the sum \(\sum_{r=1}^n \frac{r^n C_r}{{ }^n C_{r-1}}\).

Solution: We know the ratio formula:
\(
\frac{{ }^n C_r}{{ }^n C_{r-1}}=\frac{n-r+1}{r}
\)
Invert the ratio
\(
\frac{r^n C_r}{{ }^n C_{r-1}}=r \cdot \frac{{ }^n C_r}{{ }^n C_{r-1}}=r \cdot \frac{n-r+1}{r}=n-r+1
\)
Sum the terms
\(
\sum_{r=1}^n(n-r+1)
\)
Notice that when \(r=1\), the term is \(n-1+1=n\) When \(r=2\), the term is \(n-2+1=n-1 \ldots\) and finally, when \(r=n\), the term is \(n-n+1=1\).
So the sum is:
\(
n+(n-1)+(n-2)+\cdots+1
\)
Use the formula for the sum of first \(n\) natural numbers
\(
1+2+\cdots+n=\frac{n(n+1)}{2}
\)
Here, the sum is in reverse order, but the sum is the same:
\(
\sum_{r=1}^n(n-r+1)=1+2+\cdots+n=\frac{n(n+1)}{2}
\)

Integral and Fractional Parts

In order to find the integral and fractional parts of irrational numbers of the form \((a+b \sqrt{c})^n\), where \(a, b, c\) and \(n\) are natural numbers, let us consider the irrational number \(R=(5 \sqrt{5}+11)^{2 n+1}\).
Let \(F\) be its fractional part i.e. \(F=R-[R]\).
Let \(G=(5 \sqrt{5}-11)^{2 n+1}\). Clearly, \(0<G<1\).
Now,
\(
\begin{aligned}
& R-G =(5 \sqrt{5}+11)^{2 n+1}-(5 \sqrt{5}-11)^{2 n+1} \\
\Rightarrow & R-G =2\left\{{ }^{2 n+1} C_1(5 \sqrt{5})^{2 n} \times(11)^1+\ldots+{ }^{2 n+1} C_3(5 \sqrt{5})^{2 n-2}(11)^3+\ldots .\right\}
\end{aligned}
\)
\(
\begin{array}{ll}
\Rightarrow & R-G=\text { An even natural number } \\
\Rightarrow & {[R]+F-G=\text { An even natural number }} \\
\Rightarrow & F-G=\text { An integer } \\
\Rightarrow & F-G=0 \\
\Rightarrow & F=G=(5 \sqrt{5}-11)^{2 n+1}
\end{array}
\)
Hence, fractional part of \((5 \sqrt{5}+11)^{2 n+1}\) is \((5 \sqrt{5}-11)^{2 n+1}\).
Now, consider the irrational number \((5+2 \sqrt{6})^n\).
Let \(I\) and \(F\) denote its integral and fractional parts. Then,
\(
(5+2 \sqrt{6})^n=I+F
\)
Let \(G=(5-2 \sqrt{6})^n\). Then,
\(
\begin{aligned}
& I+F+G=(5+2 \sqrt{6})^n+(5-2 \sqrt{6})^n \\
\Rightarrow & I+F+G=2\left({ }^n C_0 5^n+{ }^n C_2 5^{n-2}(2 \sqrt{6})^2+\ldots . .\right) \\
\Rightarrow & I+F+G=\text { An even integer } \\
\therefore & F+G \text { is an integer } \\
\Rightarrow & F+G=1 \\
\Rightarrow & F=1-G
\end{aligned}
\)
Hence, fractional part of \((5+2 \sqrt{6})^n\) is \(1-(5-2 \sqrt{6})^n\). The above discussion suggests us the following algorithm to find the integral and fractional parts of an irrational number of the form \((a+b \sqrt{c})^n\).

General algorithm for \((a+b \sqrt{c})^n\)

Let \(R=(a+b \sqrt{c})^n, G=(a-b \sqrt{c})^n\)
If \(G<1\), then:
(i) If \(n\) is odd, fractional part \(F=G\)
(ii) If \(n\) is even, fractional part \(F=1-G\)
Integral part is then \(I=R-F\)
This method works for any natural numbers \(a, b, c, n\).

For example: \(R=(5+2 \sqrt{6})^n\)
Let \(F=\) fractional part, \(I=\) integer part
Let \(G=(5-2 \sqrt{6})^n\)
We know:
\(
I+F+G=(5+2 \sqrt{6})^n+(5-2 \sqrt{6})^n=\text { an integer. }
\)
Also, \(0<G<1\), so \(F+G=1 \Longrightarrow F=1-G\)
Hence, the fractional part is:
\(
F=1-(5-2 \sqrt{6})^n
\)

Algorithm

STEP I: Write the given expression equal to \(I+F\), where \(I\) is its integral part and \(F\) is the fractional part.
STEP II: Define \(G\) by replacing ‘ + ‘ sign in the given expression by ‘ – ‘. Note that \(G\) always lies between 0 and 1.
STEP III: Either add \(G\) to the expression in step I or subtract \(G\) from the expression in step I so that RHS is an integer.
STEP IV: If \(G\) is added to the expression in step I, then \(G+F\) will always come out to be equal to 1 i.e. \(G=1-F\). If \(G\) is subtracted from the expression in step \(I\), then \(G\) will always come out to be equal to \(F\).
STEP V: Obtain the value of the desired expression after getting \(F\) in terms of \(G\).

Remark: For a number of the form \(G=(a-b \sqrt{c})^n\) :
If \(a<b \sqrt{c}\), then \(0<a-b \sqrt{c}<1\)
Raising a number in \((0,1)\) to any positive power gives a number in \((0,1)\)
This is why \(G\) is always between 0 and 1 in such problems.

For example: \(G=(5-2 \sqrt{6})^n\)
Compute \(5-2 \sqrt{6}\) :
\(
\begin{aligned}
& \sqrt{6} \approx 2.4495 \\
& \Rightarrow 2 \sqrt{6} \approx 4.899 \\
& \Rightarrow 5-2 \sqrt{6} \approx 0.101
\end{aligned}
\)
Again, \(0<5-2 \sqrt{6}<1\), so raising to any positive integer \(n\) gives:
\(
0<G<1
\)

Example 20: The greatest integer less than or equal to \((\sqrt{2}+1)^6\) is
(a) 197
(b) 198
(c) 196
(d) 199

Solution: (a) Let \((\sqrt{2}+1)^6=I+F\), where \(I \in N\) and \(0<F<1\). Clearly, \(I\) is the greatest integer less than or equal to \((\sqrt{2}+1)^6\).
Let \(G=(\sqrt{2}-1)^6\). Then,
\(
\begin{aligned}
& I+F+G=(\sqrt{2}+1)^6+(\sqrt{2}-1)^6 \\
\Rightarrow & I+F+G=2\left\{{ }^6 C_0(\sqrt{2})^6+{ }^6 C_2(\sqrt{2})^{6-2}+\ldots\right\} \\
\Rightarrow & I+F+G=\text { An even integer } \\
\Rightarrow & F+G=1
\end{aligned}
\)
\(
\begin{aligned}
& \text { Again, } I+F+G \\
& \qquad=2\left\{{ }^6 C_0(\sqrt{2})^6+{ }^6 C_2(\sqrt{2})^{6-2}+{ }^6 C_4(\sqrt{2})^{6-4}+{ }^6 C_6(\sqrt{2})^{6-6}\right\}
\end{aligned}
\)
\(
\begin{aligned}
\Rightarrow & I+1 =2(8+15 \times 4+15 \times 2+1) & {[\because F+G=1] } \\
\Rightarrow & I =197 &
\end{aligned}
\)

Example 21: If \(R=(6 \sqrt{6}+14)^{2 n+1}\) and \(f=R-[R]\), where [ ] denotes the greatest integer function, then \(R f\) equals
(a) \(20^n\)
(b) \(20^{2 n}\)
(c) \(20^{2 n+1}\)
(d) none of these

Solution: (c) Let \(G=(6 \sqrt{6}-14)^{2 n+1}\). Then,
\(
\begin{array}{ll}
& R-G=(6 \sqrt{6}+14)^{2 n+1}-(6 \sqrt{6}-14)^{2 n+1} \\
\Rightarrow & R-G=2\left\{{ }^{2 n+1} C_1(6 \sqrt{6})^{2 n}(14)+\ldots\right\} \\
\Rightarrow & R-G=\text { An even integer } \\
\Rightarrow & {[R]+f-G=\text { An even integer }} \\
\Rightarrow & f=G \\
\therefore & R f=R G=(6 \sqrt{6}+14)^{2 n+1}(6 \sqrt{6}-14)^{2 n+1}=20^{2 n+1}
\end{array}
\)

Explanation:
Step 1: Use the conjugate
Let the conjugate be:
\(
G=\bar{R}=(6 \sqrt{6}-14)^{2 n+1} .
\)
Since \(6 \sqrt{6} \approx 14.6969\), we have:
\(
0<14-6 \sqrt{6}=-(6 \sqrt{6}-14)<1
\)
Therefore, \(\bar{R}=(6 \sqrt{6}-14)^{2 n+1}\) satisfies:
\(
0<f=R-[R]=\bar{R}<1
\)
So fractional part:
\(
f=(6 \sqrt{6}-14)^{2 n+1}
\)
Step 2: Express \(R f\)
\(
R f=R \cdot(6 \sqrt{6}-14)^{2 n+1}=((6 \sqrt{6}+14)(6 \sqrt{6}-14))^{2 n+1}
\)
Step 3: Multiply conjugates
\(
(6 \sqrt{6}+14)(6 \sqrt{6}-14)=(6 \sqrt{6})^2-14^2=36 \cdot 6-196=216-196=20
\)
Step 4: Raise to the power
\(
R f=20^{2 n+1}
\)

Example 22: If \(R=(\sqrt{2}+1)^{2 n+1}\) and \(f=R-[R]\), where [ ] denotes the greatest integer function, then \([R]\) equal
(a) \(f+\frac{1}{f}\)
(b) \(f-\frac{1}{f}\)
(c) \(\frac{1}{f}-f\)
(d) none of these

Solution: (c) Let \(G=(\sqrt{2}-1)^{2 n+1}\). Then,
\(
\begin{aligned}
& R-G=(\sqrt{2}+1)^{2 n+1}-(\sqrt{2}-1)^{2 n+1} \\
\Rightarrow & R-G=2\left\{{ }^{2 n+1} C_1(\sqrt{2})^{2 n}+{ }^{2 n+1} C_3(\sqrt{2})^{2 n-2}+\ldots\right\} \\
\Rightarrow & R-G=\text { An even integer } \\
\Rightarrow & {[R]+f-G=\text { An even integer }=I \text { (say) } } \\
\Rightarrow & f-G=0 \Rightarrow f=G
\end{aligned}
\)
Now,
\(
\begin{aligned}
& f=G \\
\Rightarrow \quad & \frac{1}{f}=\frac{1}{G}=\frac{1}{(\sqrt{2}-1)^{2 n-1}}=(\sqrt{2}+1)^{2 n+1}=R \\
\Rightarrow \quad & \frac{1}{f}=R \Rightarrow \frac{1}{f}=f+[R] \Rightarrow[R]=\frac{1}{f}-f
\end{aligned}
\)

Example 23: If \((5+2 \sqrt{6})^n=I+f\), where \(I \in N, n \in N\) and \(0 \leq f<1\), then I equals
(a) \(\frac{1}{f}-f\)
(b) \(\frac{1}{1+f}-f\)
(c) \(\frac{1}{1-f}-f\)
(d) \(\frac{1}{1-f}+f\)

Solution: (c) Let \(G=(5-2 \sqrt{6})^n\). Then,
\(
\begin{aligned}
& I+f+G=(5+2 \sqrt{6})^n+(5-2 \sqrt{6})^n \\
\Rightarrow & I+f+G=2\left\{{ }^n C_0 5^n(2 \sqrt{6})^0+{ }^n C_2 5^{n-2}(2 \sqrt{6})^2+\ldots\right\} \\
\Rightarrow & I+f+G=\text { An even integer } \\
\Rightarrow & f+G=1
\end{aligned}
\)
Now,
\(
\begin{aligned}
& I+f=(5+2 \sqrt{6})^n \\
\Rightarrow \quad & I=(5+2 \sqrt{6})^n-f \\
\Rightarrow \quad & I=\frac{1}{(5-2 \sqrt{6})^n}-f
\end{aligned}
\)
\(
\Rightarrow \quad I=\frac{1}{1-f}-f \quad\left[\begin{array}{l}
\left.\because(5-2 \sqrt{6})^n=G \text { and } f+G=1\right] \\
\therefore(5-2 \sqrt{6})^n=1-f
\end{array}\right]
\)

Example 24: If \(R=(7+4 \sqrt{3})^{2 n}=I+f\), where \(I \in N\) and \(0<f<1\), then \(R(1-f)\) equals
(a) \((7-4 \sqrt{3})^{2 n}\)
(b) \(\frac{1}{(7+4 \sqrt{3})^{2 n}}\)
(c) 1
(d) none of these

Solution: (c) Let \(G=(7-4 \sqrt{3})^{2 n}\). Then,
\(
\begin{array}{ll}
& R+G=(7+4 \sqrt{3})^{2 n}+(7-4 \sqrt{3})^{2 n} \\
\Rightarrow & R+G=2\left\{{ }^{2 n} C_0 7^{2 n}+{ }^{2 n} C_2 7^{2 n-2}(4 \sqrt{3})^2+\ldots\right\} \\
\Rightarrow & R+G=\text { An even integer } \\
\Rightarrow & I+f+G=\text { An even integer } \\
\Rightarrow & f+G=\text { An integer } \\
\Rightarrow & f+G=1[\because 0 \leq f<1,0<G<1 \therefore 0<f+G<2 \Rightarrow f+G=1] \\
\Rightarrow & G=1-f \\
\therefore & R(1-f)=R G=(7+4 \sqrt{3})^{2 n}(7-4 \sqrt{3})^{2 n}=1
\end{array}
\)

Greatest Term in Binomial Expansion

The Binomial Theorem provides a formula to expand any power of a binomial expression \((a+b)^n\). The general term (or \((k+1)\)-th term, where \(k\) starts from 0 ) of the expansion is given by the formula:
\(
T_{k+1}=\binom{n}{k} a^{n-k} b^k
\)
where \(\binom{n}{k}=\frac{n!}{k!(n-k)!}\) is the binomial coefficient.
For the given expansion of \((1+x)^7\), we have \(a=1, b=x\), and \(n=7\). When evaluated at \(x=1 / 2\), the formula for the terms becomes:
\(
T_{k+1}=\binom{7}{k}(1)^{7-k}(1 / 2)^k=\binom{7}{k}\left(\frac{1}{2}\right)^k
\)
We calculate the coefficients \(\binom{7}{k}\) for \(k=0\) through \(k=7\) and then multiply by the corresponding power of 1/2.
Calculating each term \(T_{k+1}\) :
\(
\begin{aligned}
& \text { Term 1 }(k=0):\binom{7}{0}(1 / 2)^0=1 \times 1=1 \\
& \text { Term 2 }(k=1):\binom{7}{1}(1 / 2)^1=7 \times 1 / 2=7 / 2 \\
& \text { Term 3 }(k=2):\binom{7}{2}(1 / 2)^2=21 \times 1 / 4=21 / 4 \\
& \text { Term 4 }(k=3):\binom{7}{3}(1 / 2)^3=35 \times 1 / 8=35 / 8 \\
\end{aligned}
\)
\(
\begin{aligned}
& \text { Term 5 }(k=4):\binom{7}{4}(1 / 2)^4=35 \times 1 / 16=35 / 16 \\
& \text { Term 6 }(k=5):\binom{7}{5}(1 / 2)^5=21 \times 1 / 32=21 / 32 \\
& \text { Term 7 }(k=6):\binom{7}{6}(1 / 2)^6=7 \times 1 / 64=7 / 64 \\
& \text { Term 8 }(k=7):\binom{7}{7}(1 / 2)^7=1 \times 1 / 128=1 / 128
\end{aligned}
\)

Here, we can observe that value of the term increases till the \(3^{\text {rd }}\) term and then the value of the term decreases. Then, here \(T_3\) is the greatest term. \(T_3\) is the greatest term, hence \(T_3 / T_2>1\) and also \(T_4 / T_3<1\).
In any binomial expansion, the values of the terms increase, reach a maximum and then decreases.

So in general to locate the maximum term in the expansion \((1+x)^n\) we find the value of \(r\) till the ratio \(T_{r+1} / T_r\) is greater than 1 as for this value of \(r\), any term is always greater than its previous term. The value of \(r\) till this occurs gives the greatest term.
Derive the greatest term:
For the expansion of
\(
(1+x)^n,
\)
the general term is
\(
T_{r+1}=\binom{n}{r} x^r.
\)
Ratio of successive terms:
\(
\frac{T_{r+1}}{T_r}=\frac{\binom{n}{r} x^r}{\binom{n}{r-1} x^{r-1}}=\frac{n-r+1}{r} x.
\)
Condition for greatest term:
The greatest term occurs when the terms increase and then decrease, so we require:
\(
\frac{T_{r+1}}{T_r} \geq 1.
\)
Thus,
\(
\frac{n-r+1}{r} x \geq 1.
\)
Solve the inequality
\(
\begin{gathered}
(n-r+1) x \geq r \\
n x+x \geq r(1+x) \\
r \leq \frac{(n+1) x}{1+x}
\end{gathered}
\)
The greatest term occurs for the value of \(r\) satisfying
\(
r \leq \frac{(n+1) x}{1+x}.
\)

Then the greatest term occurs for \(r=[(n+1) x /(1+x)]\), which is the integral part of \([(n+1) x /(1+x)]\). If \([(n+1) x /(1+x)]\) is an exact integer, then \(T_r\) and \(T_{r+1}\) both are the greatest terms.

Case-I: For example : Single greatest term (non-integer case)
Find the greatest term in the expansion of
\(
(1+2)^7
\)
Here,
\(
\begin{aligned}
n & =7, \quad x=2. \\
\frac{(n+1) x}{1+x} & =\frac{8 \cdot 2}{3}=\frac{16}{3}=5.33 \ldots
\end{aligned}
\)
So,
\(
r=\lfloor 5.33 \ldots\rfloor=5.
\)
Hence the greatest term is:
\(
\begin{gathered}
T_{r+1}=T_6 \\
T_6=\binom{7}{5} 2^5=21 \times 32=672
\end{gathered}
\)

Case-II: For example: Two greatest terms (exact integer case)
Find the greatest terms in the expansion of
\(
(1+3)^3
\)
Here,
\(
\begin{gathered}
n=3, \quad x=3 . \\
\frac{(n+1) x}{1+x}=\frac{4 \cdot 3}{4}=3 \quad(\text { exact integer })
\end{gathered}
\)
So the two greatest terms are:
\(
T_3 \text { and } T_4.
\)
Compute them:
\(
\begin{aligned}
& T_3=\binom{3}{2} 3^2=3 \times 9=27 \\
& T_4=\binom{3}{3} 3^3=1 \times 27=27
\end{aligned}
\)

When we have an expansion in which positive and negative sign occurs alternatively, we find the numerically greatest term (ignoring the negative value), for which we find the value of \(r\) considering ratio \(\left|T_{r+1} / T_r\right|\).

Note: The greatest coefficient in the binomial expansion is equivalent to the greatest term when \(x=1\).

Above discussion suggests the following algorithm to find the greatest term in a binomial expansion.

Algorithm

STEP I: Write \(T_{r+1}\) and \(T_r\) from the given expansion.
STEP II: Find \(\frac{T_{r+1}}{T_r}\)
STEP III: Put \(\frac{T_{r+1}}{T_r}>1\)
STEP IV: Solve the inequality in step III for \(r\) to get an inequality of the form \(r<m\) or \(r>m\).
If \(m\) is an integer, then \(m\) th and \((m+1)\) th terms are equal in magnitude and these two are the greatest terms. If \(m\) is not an integer, then obtain the integral part of \(m\), say \(k\). In this case, \((k+1)^{\text {th }}\) term is the greatest term.

Example 25: The greatest term in the expansion of \((1+x)^{10}\) when \(x=2 / 3\), is
(a) \(4^{\text {th }}\)
(b) \(5^{\text {th }}\)
(c) \(6^{\text {th }}\)
(d) \(3^{\text {rd }}\)

Solution: (b) Let \(T_r\) and \(T_{r+1}\) denote the \(r^{\text {th }}\) and \((r+1)^{\text {th }}\) terms in the expansion of \((1+x)^{10}\). Then,
\(
\begin{array}{rlrl}
T_r & ={ }^{10} \mathrm{C}_{r-1} x^{r-1} \text { and } T_{r+1}={ }^{10} \mathrm{C}_r x^r . \\
\therefore & \frac{T_{r+1}}{T_r} & =\frac{{ }^{10} \mathrm{C}_r x^r}{{ }^{10} \mathrm{C}_{r-1} x^{r-1}} \\
\Rightarrow \quad \frac{T_{r+1}}{T_r} & =\frac{{ }^{10} \mathrm{C}_r}{{ }^{10} \mathrm{C}_{r-1}} x
\end{array}
\)
\(
\begin{aligned}
& \Rightarrow \quad \frac{T_{r+1}}{T_r}=\frac{10!}{(10-r)!r!} \times \frac{(10-r+1)!(r-1)!}{10!} x \\
& \Rightarrow \quad \frac{T_{r+1}}{T_r}=\frac{11-r}{r} \cdot x \Rightarrow \frac{T_{r+1}}{T_r}=\left(\frac{11-r}{r}\right) \times \frac{2}{3} \quad[\because x=2 / 3]
\end{aligned}
\)
Now,
\(
\frac{T_{r+1}}{T_r}>1 \Rightarrow\left(\frac{11-r}{r}\right) \times \frac{2}{3}>1 \Rightarrow 22>5 r \Rightarrow r<4 \frac{2}{5}
\)
\(\therefore \quad(4+1)^{\text {th }}\) i.e. 5 th term is the greatest term.

Example 26: The greatest term in the expansion of \((1+3 x)^{54}\) when \(x=\frac{1}{3}\), is
(a) \(28^{\text {th }}\)
(b) \(25^{\text {th }}\)
(c) \(26^{\text {th }}\)
(d) \(24^{\text {th }}\)

Solution: (a) Let \(T_{r+1}\) and \(T_r\) denote the \((r+1)^{\text {th }}\) and \(r^{\text {th }}\) terms respectively. Then,
\(
\begin{aligned}
& \quad T_{r+1}={ }^{54} \mathrm{C}_r(3 x)^r \text { and } T_r={ }^{54} \mathrm{C}_{r-1}(3 x)^{r-1} \\
& \therefore \quad \frac{T_{r+1}}{T_r}=\frac{{ }^{54} \mathrm{C}_r}{{ }^{54} \mathrm{C}_{r-1}} \times(3 x)^r=\frac{54-r+1}{r} \times(3 x)^r \\
& \Rightarrow \quad \frac{T_{r+1}}{T_r}=\frac{55-r}{r} \quad\left[\because x=\frac{1}{3}\right] \\
& \text { Now }, \frac{T_{r+1}}{T_r}>1 \Rightarrow \frac{55-r}{r}>1 \Rightarrow r<27 \frac{1}{2}
\end{aligned}
\)
Hence, 28th term is the greatest term.

Example 27: Find the greatest term in the expansion of \(\sqrt{3}\left(1+\frac{1}{\sqrt{3}}\right)^{20}\).

Solution: Let \((r+1)^{\text {th }}\) term be the greatest term in the expansion of
\(
\begin{aligned}
& \left(1+\frac{1}{\sqrt{3}}\right)^{20} \cdot \text { Now, } \\
& \frac{T_{r+1}}{T_r}=\frac{20-r+1}{r}\left(\frac{1}{\sqrt{3}}\right)
\end{aligned}
\)
Let
\(
\begin{aligned}
& T_{r+1} \geq T_r \\
\Rightarrow & 20-r+1 \geq \sqrt{3} r \\
\Rightarrow & 21 \geq r(\sqrt{3}+1)
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow r \leq \frac{21}{\sqrt{3}+1} \\
& \Rightarrow r \leq 7.686 \\
& \Rightarrow r=7
\end{aligned}
\)
for which the greatest term occurs.
Hence, the greatest term in \(\sqrt{3}\left(1+\frac{1}{\sqrt{3}}\right)^{20}\) is
\(
T_8=\sqrt{3}{ }^{20} C_7\left(\frac{1}{\sqrt{3}}\right)^7=\frac{25840}{9}
\)

Example 28: Find the numerically greatest term in the expansion of \((3-5 x)^{15}\) when \(x=1 / 5\).

Solution: \((3-5 x)^{15}=3^{15}(1-5 x / 3)^{15}=3^{15}(1-1 / 3)^{15}(\) for \(x=1 / 5)\)
Now consider \((1-1 / 3)^{15}\).
\(
\begin{aligned}
& \left|\frac{T_{r+1}}{T_r}\right|=\frac{15-r+1}{r}\left|-\frac{1}{3}\right| \geq 1 \\
\Rightarrow & 16-r \geq 3 r \\
\Rightarrow & r \leq 4
\end{aligned}
\)
Hence, \(T_4\) and \(T_5\) are the numerically greatest terms.
\(
T_4={ }^{15} C_3 3^{15-3}(-5 x)^3=-455 \times\left(3^{12}\right)
\)
and
\(
T_5={ }^{15} C_4 3^{15-4}(-5 x)^4=455 \times\left(3^{12}\right)
\)
Also, \(\left|T_4\right|=\left|T_5\right|=455 \times\left(3^{12}\right)\).

Example 29: Given that the \(\mathbf{4}^{\text {th }}\) term in the expansion of \([2+(3 / 8) x]^{10}\) has the maximum numerical value. Then find the range of value of \(x\).

Solution: Let \(T_4\) be numerically the greatest term in the expansion of
\(
\begin{gathered}
2^{10}\left[1+\left(\frac{3}{16}\right) x\right]^{10} . \text { Therefore, } \\
\left|\frac{T_4}{T_3}\right| \geq 1 \text { and }\left|\frac{T_4}{T_5}\right| \geq 1 \\
\Rightarrow\left|\frac{T_4}{T_3}\right| \geq 1 \text { and }\left|\frac{T_5}{T_4}\right| \leq 1
\end{gathered}
\)
Now,
\(
\frac{T_{r+1}}{T_r}=\frac{n-r+1}{r} x
\)
Taking \(r=3\) and \(r=4\) and replacing \(x\) by \(3 x / 16\) and putting \(n=10\) in the above two relations, we get
\(
\begin{aligned}
& \left|\frac{11-3}{3} \frac{3 x}{16}\right| \geq 1 \text { and }\left|\frac{11-4}{4} \frac{3 x}{16}\right| \leq 1 \\
\Rightarrow & |x| \geq 2 \text { and }|x| \leq \frac{64}{21} \dots(1)
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \quad x^2 \geq 4 \text { and } x^2 \leq\left(\frac{64}{21}\right)^2 \\
& \Rightarrow \quad 4 \leq x^2 \leq\left(\frac{64}{21}\right)^2 \\
& \Rightarrow \quad 2 \leq x \leq \frac{64}{21} \text { or }-\frac{64}{21} \leq x \leq-2
\end{aligned}
\)

Example 30: Find the greatest coefficient in the expansion of \((1+2 x / 3)^{15}\).

Solution: The greatest coefficient is equal to the greatest term when \(x=1\).
For \(x=1, \frac{T_{r+1}}{T_r}=\frac{15-r+1}{r} \frac{2}{3}\)
\(
\begin{aligned}
& \text { Let } \frac{T_{r+1}}{T_r} \geq 1 \\
& \Rightarrow \frac{15-r+1}{r} \frac{2}{3} \geq 1 \\
& \Rightarrow 32-2 r \geq 3 r \\
& \Rightarrow r \leq 32 / 5 \\
& \Rightarrow r=6
\end{aligned}
\)
Hence, \(7^{\text {th }}\) term has the greatest coefficient and its value is \(T_{6+1}={ }^{15} C_6(2 / 3)^6\).