General and Middle Terms - Practice Questions & MCQ

General Term

Sometimes we are interested only in a certain term of a binomial expansion. We do not need to fully expand a binomial to find a single specific term. 

$(r+1)^{\text {th }}$ term is called as general term in $(x+y)^n$ and general term is given by

$
\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{n}-\mathrm{r}} \cdot \mathrm{y}^{\mathrm{r}}
$
Term independent of $\mathbf{x}$
It means a term containing $x^0$,
For example, to find term independent of x in $\left(x-\frac{1}{x}\right)^{20}$

$
\begin{aligned}
& \left(x-\frac{1}{x}\right)^{20} \\
& \Rightarrow \quad T_{r+1}={ }^{20} C_r x^{20-r}(-1)^r \frac{1}{x^r}=20 C_r x^{20-2 r}(-1)^r \\
& \Rightarrow \quad 20-2 r=0 ; r=10 \\
& \Rightarrow \quad 11^{\text {th }} \text { term is independent of } x
\end{aligned}
$

$(p+1)^{\text {th }}$ term from the end
The binomial expansion

$
(\mathrm{x}+\mathrm{y})^{\mathrm{n}}={ }^{\mathrm{n}} \mathrm{C}_0 \mathrm{x}^{\mathrm{n}}+{ }^{\mathrm{n}} \mathrm{C}_1 \mathrm{x}^{\mathrm{n}-1} \mathrm{y}+{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^{\mathrm{n}-2} \mathrm{y}^2+\cdots+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{y}^{\mathrm{n}}
$

From Starting

$
\underbrace{{ }^n \mathrm{C}_0 x^n}_{\begin{array}{c}
1^{\text {st }} \\
\text { term }
\end{array}}+\underbrace{{ }^n \mathrm{C}_1 x^{n-1} y}_{\begin{array}{c}
2^{\text {nd }} \\
\text { term }
\end{array}}+\underbrace{{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^{\mathrm{n}-2} y^2}_{\begin{array}{c}
\mathbf{n}^{\text {th }} \\
\text { term }
\end{array}}+\cdots+\underbrace{{ }^n C_{n-1} x y_n-1}_{\begin{array}{c}
(\mathrm{n}+1)^{\mathrm{th}} \\
\text { term }
\end{array}} \underbrace{n}
$

From the End

$
\underbrace{{ }^n \mathrm{C}_0 y^n}_{\begin{array}{c}
1^{\text {st }} \\
\text { term }
\end{array}}+\underbrace{{ }^n \mathrm{C}_1 y^{n-1} x}_{\begin{array}{c}
2^{\text {nd }} \\
\text { term }
\end{array}}+{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{y}^{\mathrm{n}-2} \mathrm{x}^2+\cdots+\underbrace{{ }^n C_{n-1} y x^{n-1}}_{\begin{array}{c}
\mathbf{n}^{\text {th }} \\
\text { term }
\end{array}}+\underbrace{{ }^n \mathrm{C}_n x^n}_{\begin{array}{c}
(\mathrm{n}+1)^{\text {th }} \\
\text { term }
\end{array}}
$

(Using relation ${ }^n C_r={ }^n C_{(n-r) \text { ) }}$
Now,
Consider the binomial expansion

$
(\mathrm{y}+\mathrm{x})^{\mathrm{n}}={ }^{\mathrm{n}} \mathrm{C}_0 \mathrm{y}^{\mathrm{n}}+{ }^{\mathrm{n}} \mathrm{C}_1 \mathrm{y}^{\mathrm{n}-1} \mathrm{x}+{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{y}^{\mathrm{n}-2} \mathrm{x}^2+\cdots+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}}
$
Just observe that, $(p+1)^{\text {th }}$ term from the end of the expansion of $(x+y)^n=(p+1)^{\text {th }}$ term from the beginning of the expansion of

$
(y+x)^n={ }^n C_p y^{n-p} x^p
$

Rational term in the expansion of $\left(\mathrm{x}^{1 / \mathrm{a}}+\mathrm{y}^{1 / \mathrm{b}}\right)^{\mathrm{N}}, \mathrm{x}, \mathrm{y} \in$ prime numbers.
First, find $T_{r+1}={ }^N C_r\left(x^{1 / a}\right)^{N-r}\left(y^{1 / b}\right)^r$

$
\therefore \quad T_{r+1}={ }^N C_r \cdot x^{(N-r) / a} \cdot y^{r / b}
$
By observation, when indices of $x$ and $y$ are integers, then the entire term will be rational.

For example,
Find the number of terms in the expansion of $(\sqrt[4]{9}+\sqrt[6]{8})^{100}$ which are rational
To make x and y as prime numbers, we can rewrite the expression as

$
\therefore \text { General term, } \begin{aligned}
T_{r+1} & ={ }^{100} C_r\left(3^{1 / 2}\right)^{100-r} \cdot\left(2^{1 / 2}\right)^r \\
& ={ }^{100} C_r \cdot 3^{\frac{100-r}{2}} \cdot 2^{r / 2} \\
& ={ }^{100} C_r \cdot 3^{50-r / 2} \cdot 2^{r / 2}
\end{aligned}
$
Now, $\quad 0 \leq r \leq 100$
For $r=0,2,4,6,8, \ldots, 100$, indices of 3 and 2 are positive integers.
Hence, the number of terms which are rational $=50+1=51$

The middle term in the expansion $(x+y)^n$, depends on the value of ' $n$ '.
Case 1 When ' $n$ ' is even
If n is even, and the number of terms in the expansion is $\mathrm{n}+1$, so $\mathrm{n}+1$ is an odd number therefore only one middle term is obtained which is

$
\left(\frac{\mathrm{n}}{2}+1\right)_{\text {term }}^{\mathrm{th}}
$
It is given by

$
\mathrm{T}_{\frac{\mathrm{n}}{2}+1}=\binom{\mathrm{n}}{\frac{\mathrm{n}}{2}} x^{\frac{\mathrm{n}}{2}} y^{\frac{\mathrm{n}}{2}}
$
Case 2 When ' $n$ ' is odd

It is given by

$
T_{\frac{n+1}{2}}=\binom{n}{\frac{n1}{2}} x^{\frac{n+1}{2}} \cdot y^{\frac{n1}{2}} \text { and } T_{\frac{n+3}{2}}=\binom{n}{\frac{n+1}{2}} x^{\frac{n1}{2}} \cdot y^{\frac{n+1}{2}}
$
Note:
The Binomial Coefficient of the middle term is the greatest among all binomial coefficients in an expansion.
So if $n$ is even, then ${ }^n C_r$ is largest if $r=n / 2$
And if n is odd, then ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}$ is largest if $\mathrm{r}=\frac{\frac{n1}{2}}{}$ or $\frac{n+1}{2}$, and both these values of ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}$ are equal.

If consecutive coefficients are given

If in the question, coefficients of consecutive terms, like $r^{\text {th }},(r+1)^{\text {th }}$ and $(r+2)^{\text {th }}$ and so on $\ldots$ are involved, then to solve divide consecutive coefficients pairwise.

$
\text { i.e. } \frac{\text { coefficient of }(\mathrm{r}+1) \text { th of term }}{\text { coefficient of }(\mathrm{r}) \text { th of term }}, \frac{\text { coefficient of }(\mathrm{r}+2) \text { th of term }}{\text { coefficient of }(\mathrm{r}+1) \text { th of term }} \ldots
$
And with the help of the given condition in the question, you get equations. Solve these equations to get the answer.
Let's understand this by an example,
If the coefficients of $\mathrm{r}^{\text {th }},(\mathrm{r}+1)^{\text {th }}$ and $(\mathrm{r}+2)^{\text {th }}$ terms of the expansion $(1+x)^{\mathrm{n}}$ are in AP, and the relation between n and r is asked (where n is a positive integer), then
From the concept of the general term

$
\begin{aligned}
T_r & =T_{(r-1)+1}={ }^n C_{r-1} x^{r-1} \\
T_{r+1} & ={ }^n C_r x^r \text { and } T_{r+2}=T_{(r+1)+1}={ }^n C_{r+1} x^{r+1}
\end{aligned}
$

$\therefore$ Coefficients of $r$ th, $(r+1)$ th and $(r+2)$ th terms in the expansion of

$
(1+x)^n \text { are }{ }^n C_{r-1},{ }^n C_r,{ }^n C_{r+1}
$

$\because$ Given, ${ }^n C_{r-1},{ }^n C_r,{ }^n C_{r+1}$ are in AP. and $\quad n \geq r+1$

\begin{aligned}
& \therefore \frac{{ }^n C_{r-1}}{{ }^n C_r}, 1, \frac{{ }^n C_{r+1}}{{ }^n C_r} \text { are also in AP. } \\
& \Rightarrow \frac{r}{n-r+1}, 1, \frac{n-r}{r+1} \text { are in AP. } \\
& \Rightarrow 1-\frac{r}{n-r+1}=\frac{n-r}{r+1}-1 \Rightarrow \frac{n-2 r+1}{n-r+1}=\frac{n-2 r-1}{r+1} \\
& \Rightarrow n r-2 r^2+r+n-2 r+1=n^2-2 n r-n-n r+2 r^2+r+n-2 r-1 \\
& \Rightarrow \quad n^2-4 n r+4 r^2=n+2 \Rightarrow(n-2 r)^2=n+2
\end{aligned}

If consecutive terms are given 

If in the question, consecutive terms $T_r, T_{r+1}, T_{r+2}$, and so on.... are given. Then divide consecutive coefficients pairwise i.e. $\frac{\mathrm{T}_{\mathrm{r}+1}}{\mathrm{~T}_{\mathrm{r}}}=\alpha_1, \frac{\mathrm{~T}_{\mathrm{r}+2}}{\mathrm{~T}_{\mathrm{r}+1}}=\alpha_2, \frac{\mathrm{~T}_{\mathrm{r}+3}}{\mathrm{~T}_{\mathrm{r}+2}}=\alpha_3 \ldots \ldots \ldots$

Now divide $a_2$ by $a_1$ and $a_3$ by $a_2$...to solve the question.