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 $(a+b)^n$. 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, $r=14$.