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 \({ }^n \mathrm{C} a^{n-r} b^r\). The \((r+1)^{t h}\) term is also called the general term of the expansion \((a+b)^n\). It is denoted by \(\mathrm{T}_{r+1}\). Thus \(\mathrm{T}_{r+1}={ }^n \mathrm{C}_r a^{n-r} b^r\).
Regarding the middle term in the expansion \((a+b)^n\), we have
(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.
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.
(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 \(\left(\frac{n+1}{2}+1\right)^{t h}\) term. So 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.
In the expansion of \(\left(x+\frac{1}{x}\right)^{2 n}\), where \(x \neq 0\), 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.
Example 1: 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 2: 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 3: 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 4: 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 5: 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 6: \(\text { Find the term independent of } x \text { in the expansion of }\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^6 \text {. }
\)
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 7: 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 8: 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 9: 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 10: 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 11: \(\text { Find the term independent of } x \text { in the expansion of }\left(\sqrt[3]{x}+\frac{1}{2 \sqrt[3]{x}}\right)^{18}, x>0 \text {. }\)
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 12: 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 13: 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, \(r=14\).
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