General and Middle terms: The binomial theorem helps us find the power of a binomial without going through the tedious multiplication process. Further, the use of the formula helps us determine the general and middle terms in the expansion of the algebraic expressions. And also, it will be used to find the coefficient of a particular term, and we can also find an independent term, i.e. the term that does not contain any variable or is purely a number.

In this article, let us learn to find the middle and general terms of the binomial expansion $(a+b{)}^n$ using some examples.

What is the Binomial Theorem?

The binomial theorem states the principle of expanding an algebraic expression of the form $(a+b{)}^n$ and expresses it as a sum of the terms involving individual exponents of the variables, $a$ and $b$

${\left(a+b\right)}^n=\sum _{k=0}^n{}^n{C}_k{a}^{n–k}{b}^k$

$={}^n{C}_0{a}^n{b}^0+{}^n{C}_1{a}^{n-1}{b}^1+\cdots +{}^n{C}_r{a}^{n-r}{b}^r+\cdots +{}^n{C}_n{a}^{n-n}{b}^n$

where,

$n\ge 0$

${}^n{C}_k$ is a constant,

Note that every term in the binomial expansion is associated with a numeric value (constant) called a coefficient.

Learn about Algebraic Expressions here

What are Binomial Coefficients?

The binomial coefficients are positive integers that occur as coefficients in the binomial expansion. It is denoted by ${}^n{C}_r$ and it is the coefficients of $x^r$ term in the expansion of the binomial $(1+x{)}^n$.

Binomial coefficients are calculated using the formula

${}^n{C}_r=\frac{n!}{r!(n-r)!}$, where $n\ge r\ge 0$

General Term of the Binomial Theorem

We know that the binomial theorem for positive integers is given by,

${\left(a+b\right)}^n=\sum _{k=0}^n{}^n{C}_k{a}^{n–k}{b}^k$

$={}^n{C}_0{a}^n{b}^0+{}^n{C}_1{a}^{n-1}{b}^1+\cdots +{}^n{C}_r{a}^{n-r}{b}^r+\cdots +{}^n{C}_n{a}^{n-n}{b}^n$

where the coefficients of the form ${}^n{C}_r$ are called binomial coefficients.

• There are $(n+1)$ terms in the expansions of $(a+b{)}^n$.
• Observe that for the successive terms of the expansion, the index of ‘$a$’ decreases by one. It is $n$ in the first term in the expansion of $(n-1)$ in the second term, and so on, ending with zero in the last term.
• Likewise, the index of $‘b^{\prime }$ increases by one, with zero in the first term in the expansion,$1$ in the second, and so on, ending with $n$ in the last term.

Now, let’s say $T_1,T_2,T_3,T_4,\dots T_{r+1}$ are the first, second, third, and the $(n+1{)}^{th}$ terms respectively in the expansion of $(a+b{)}^n$

Now,

First term is $T_1={}^n{C}_0{a}^n$

Second term is $T_2={}^n{C}_1{a}^{n-1}b$

Third term is $T_3={}^n{C}_2\cdot a^{n-2}{b}^2$

… … … ..

The $(r+1{)}^{\text{th }}$ term is $T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$

We have the formula for the general term is

$T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$

where $0\le r\le n$

In other words, the general term is $(r+1{)}^{th}$ term of the expansion of $(a+b{)}^n$ and it is denoted by $T_{r+1}.$

This formula is used to find some specific terms, such as the term independent of $a$ or $b$ in the binomial expansion of $(a+b{)}^n$.

Independent Term of the Binomial Expansion

The independent term is a term that is independent of any variable, i.e. which gives a numerical value. It is the constant in the expansion.

The independent term can be determined by using the general term formula

$T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$

Example: Find term independent of $x$ in the following expansion ${(\frac{3}{2}{x}^2-\frac{1}{3x})}^6$

Solution: Given: Expansion $={(\frac{3}{2}{x}^2-\frac{1}{3x})}^6$

As we know, the general formula $T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$

$={}^6{C}_r{\left(\frac{3x^2}{2}\right)}^{6-r}{\left(\frac{-1}{3x}\right)}^r$

$={}^6{C}_r{\left(\frac{3}{2}\right)}^{6-r}{\left(\frac{-1}{3}\right)}^r\left(x^{12-2r-r}\right)$

We need to find the term independent of $x$ i.e the power of $x$ is zero.

Therefore, $x^{12-3r}=x^0$

$\Rightarrow 12-3r=0$

$\Rightarrow 12=3r$

$\Rightarrow r=4$

Thus, the independent term is $T_{4+1}={}^6{C}_4{\left(\frac{3}{2}\right)}^{6-4}{\left(\frac{-1}{3}\right)}^4$

$=\frac{6!}{2!4!}{\left(\frac{3^2}{2^2}\right)}^2\left(\frac{1}{3^2}\right)$

$=\frac{6\times 5\times 4!}{2!\times 4!}\times \frac{1}{2^2\times 3^2}$

Hence, the term independent of $x$ is $\frac{5}{12}$

Middle Term of the Binomial Expansion

As we know, the binomial expansion of $(a+b{)}^n$ has $(n+1)$ terms. So based on the value of $n$, we can calculate the middle term or any term of the expansion of $(a+b{)}^n$.

Case 1: $n$ is even

Since $n$ is even ,so $n+1$ is odd. Therefore, the ${\left(\frac{n+2}{2}\right)}^{\mathrm{th}}$ term is the middle term, i.e. ${(\frac{n}{2}+1)}^{\text{th }}$ or $T_{\frac{n}{2}}+1$

Case 2: $n$ is odd

If $n$ is odd, then $n+1$ is even. So there will be two middle terms that is the ${\left(\frac{n+1}{2}\right)}^{th}$ term and the ${\left(\frac{n+3}{2}\right)}^{\text{th }}$ term.

Thus, $T_{\left(\frac{n+1}{2}\right)}$ and $T_{\left(\frac{n+3}{2}\right)}$ are the two middle terms of the expansion of $(a+b{)}^n$ when $n$ is odd.

Let us make a chart for the middle terms based on the value of $n$.

Solved Examples – General and Middle Terms

Q.1. Find the $7^{\text{th }}$ term in the expansions of $(x+2{)}^{10}$
Sol: As we know, the general term in the expansion of $(a+b{)}^n$ using the binomial theorem is $T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$
Here, $r=6,n=10,a=2$
Therefore, $T_7=T_{6+1}={}^{10}{C}_6{x}^{10-6}{2}^6$
$=210x^4\times 64$ (Using ${}^n{C}_r=\frac{n!}{r!\left(n–r\right)!}$)
$\therefore T_7=13440x^4$
Hence, the $7^{\text{th }}$ term is $13440x^4$

Q.2. Find the middle term(s) in the expansions of $(x+3{)}^8$
Sol: Given thaat the expansion is $(x+3{)}^8$
Here,
$a=x$
$b=3$
$n=8$
Here, $n$ is even. Hence, there will be one middle term
$(\frac{n}{2}+1)=\frac{8}{2}+1=4+1=5$
Then, $T_5=T_{4+1}={}^8{C}_4(x{)}^{8-4}$ [Using the general term formula $T_{r+1}={}^n{C}_r{a}^{n–r}{b}^r]$
$=70x^4\times 81$ (Using ${}^n{C}_r=\frac{n!}{r!\left(n–r\right)!}$)
$=(70\times 81)x^4$
$=5670x^4$
Therefore, $5670x^4$ is the required middle term of the expansion of $(x+3{)}^8$.

Q.3. Find the middle term(s) in the expansion of $(x+2y{)}^9$
Sol: Given $(x+2y{)}^9$
Here,
$a=x$
$b=2y$
$n=9$
Since $n$ is odd, there will be two middle terms, i.e
$\frac{(n+1)}{2}=\frac{(9+1)}{2}=\frac{10}{2}=5$
And
$\frac{(n+3)}{2}=\frac{(9+3)}{2}=\frac{12}{2}=6$
Hence, $5^{\text{th }}$ and $6^{\text{th }}$ terms are the two middle terms.
Using the general term formula, $T_{r+1}={}^n{C}_r{a}^{n–r}{b}^r$
Then,
$T_5=T_{4+1}={}^9{C}_4(x{)}^{9-4}(2y{)}^4$
Using ${}^n{C}_r=\frac{n!}{r!(n-r)!}$
$=126x^5(2{)}^4(y{)}^4$
$=(126\times 16)x^5{y}^4$
$\therefore T_5=2016x^5{y}^4$
Similarly, $T_6=T_{5+1}={}^9{C}_5(x{)}^{9-5}(2y{)}^5$
$=126x^4(2{)}^5(y{)}^5$
$=(126\times 32)x^4{y}^5$
$\therefore T_6=4032x^4{y}^5$
Therefore, $2016x^5{y}^4$ and $4032x^4{y}^5$ are the required two middle terms of the expansion of $(x+2y{)}^9$.

Q.4. Find the coefficient of the independent term of $x$ in the expansion of ${(3x-\frac{2}{x^2})}^{15}$
Sol: As we know, the general term in the expansion of $(a+b{)}^n$ using the binomial theorem is $T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$
Therefore, $T_{r+1}={}^{15}{C}_r(3x{)}^{15-r}{(-\frac{2}{x^2})}^r$
$\Rightarrow T_{r+1}={}^{15}{C}_r(3{)}^{15-r}(-2{)}^r(x{)}^{15-r}{x}^{-2r}$
$\Rightarrow T_{r+1}={}^{15}{C}_r(3{)}^{15-r}(-2{)}^r(x{)}^{15-2r-r}$
Since it is independent of $x$, then $15-r-2r=0$
$\Rightarrow r=5$
$\therefore T_6={}^{15}{C}_5(3{)}^{10}(-2{)}^5$
$=-{}^{15}{C}_5(3{)}^{10}{2}^5$

Q.5. Determine the middle term(s) in the expansion of ${(\frac{x}{2}+3y)}^9$
Sol: Given ${(\frac{x}{2}+3y)}^9$
Comparing with $(a+b{)}^n$, we get $a=\frac{x}{2},b=3y,$ and $n=9$
Since $n$ is odd, there will be two middle terms
$\frac{(n+1)}{2}=\frac{(9+1)}{2}=\frac{10}{2}=5$
And
$\frac{(n+3)}{2}=\frac{(9+3)}{2}=\frac{12}{2}=6$
Thus, the $5^{\text{th }}$ and $6^{\text{th }}$ terms are the required two middle terms.
Using the general term formula, $T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$
$T_5=T_{4+1}={}^9{C}_4{\left(\frac{x}{2}\right)}^{9-4}(3y{)}^4$
$=\frac{9.8.7.6}{1.2.3.4}\left(\frac{x^5}{32}\right)\left(81y^4\right)$ (using ${}^n{C}_r=\frac{n!}{r!(n-r)!}$ )
$=\frac{5103}{16}{x}^5{y}^4$
Similarly, $T_6=T_{5+1}={}^9{C}_5{\left(\frac{x}{2}\right)}^{9-5}(3y{)}^5$
$=\frac{9.8.7.6.5}{1.2.3.4.5}\left(\frac{x^4}{16}\right)\left(243y^5\right)$
$=\frac{15309}{8}{x}^4{y}^5$
Therefore, $\frac{5103}{16}{x}^5{y}^4$ and $\frac{15309}{8}{x}^4{y}^5$ are the required two middle terms of the expansion of ${(\frac{x}{2}+3y)}^9$

Q.6. Find the $6^{\text{th }}$ term of the expansion ${(2x^2-\frac{1}{3x^2})}^{10}$
Sol:
Given: ${(2x^2-\frac{1}{3x^2})}^{10}$
Applying general term formula of $(x+a{)}^n$ is $T_{r+1}={}^n{C}_r{x}^{n-r}{a}^r$
Therefore, $T_6={}^{10}{C}_5{\left(2x^2\right)}^5{(-\frac{1}{3x^2})}^5$
$=-\frac{10!}{5!5!}\times 32\times \frac{1}{243}$
$=-\frac{896}{27}$
Hence, the $6^{\text{th }}$ term is $-\frac{896}{27}$.

Q.7. If the ratio of the coefficient of the third and fourth term in the expansion is ${(x-\frac{1}{2x})}^n$ is $1:2$, then find the value of $n$
Sol:
Given: ${(x-\frac{1}{2x})}^n$
Applying general term formula of $(x+a{)}^n$ is $T_{r+1}={}^n{C}_r{x}^{n-r}{a}^r$
Therefore, $T_3={}^n{C}_2(x{)}^{n-2}{(-\frac{1}{2x})}^2$ and $T_4={}^n{C}_3(x{)}^{n-3}{(-\frac{1}{2x})}^3$
But according to the question, we have $\frac{T_3}{T_4}=\frac{1}{2}$
$\Rightarrow \frac{{}^n{C}_2(x{)}^{n-2}{(-\frac{1}{2x})}^2}{{}^n{C}_3(x{)}^{n-3}{(-\frac{1}{2x})}^3}=\frac{1}{2}$
$\Rightarrow \frac{\frac{n!}{2!(n-2)!}(x{)}^{n-2}{(-\frac{1}{2x})}^2}{\frac{n!}{3!(n-3)!}(x{)}^{n-3}{(-\frac{1}{2x})}^3}=\frac{1}{2}$
$\Rightarrow \frac{\frac{n(n-1)(n-2)!}{2!(n-2)!}(x{)}^{n-2}{(-\frac{1}{2x})}^2}{\frac{n(n-1)(n-2)(n-3)!}{3!(n-3)!}(x{)}^{n-3}{(-\frac{1}{2x})}^3}=\frac{1}{2}$
$\Rightarrow \frac{-n(n-1)\times 3\times 2\times 1\times 8}{n(n-1)(n-2)\times 2\times 1\times 4}=\frac{1}{2}$
$\Rightarrow n-2=-12$
$\Rightarrow n=-10$
Hence, the value of $n$ is $-10$

Summary

The binomial theorem is used to expand any power of a binomial in the form of a series. The binomial theorem formula is ${\left(a+b\right)}^n=\sum _{r=0}^n\left({}^n{C}_r\right)a^{n–r}{b}^r$ where $n$ is a positive integer $a,b$ are real numbers, and $0<r\le n$. The binomial theorem is used to find the general and middle terms of the given binomial expansion. There are two cases when it comes to finding the middle term: $n$ is even, and $n$ is odd. When $n$ is even, the expansion has one middle term, and while $n$ is odd, there are two. the independent term of $x$ can also be calculated using the general term formula, i.e., $T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$.

Important Questions on General and Middle Terms

Frequently Asked Questions (FAQs)

Q.1. How do you find the middle term of expansion?
Ans: For the binomial expansion $(a+b{)}^n$, we can find middle terms as,
Since $n$ is even, so $n+1$ is odd. Therefore, the ${\left(\frac{n+1+1}{2}\right)}^{\text{th }}$ the term is the middle term, i.e ${(\frac{n}{2}+1)}^{\text{th }}$ or $T\frac{n}{2}+1$
Since $n$ odd, so $n+1$ is even. So there will be two middle terms ${\left(\frac{n+1}{2}\right)}^{\text{th }}$ term and ${(\frac{n+1}{2}+1)}^{\mathrm{th}}$ or ${\left(\frac{n+3}{2}\right)}^{\text{th }}$ term.

Q.2. What is the general term in binomial?
Ans: The general term of the binomial expansion of $(x+y{)}^n$ can be determined by using $T_{r+1}={}^n{C}_r{x}^{n-r}{y}^r$.

Q.3. What is the middle term if $n$ is odd?
Ans: If $n$ is odd, so $n+1$ is even . So there will be two middle terms, i.e., ${\left(\frac{n+1}{2}\right)}^{th}$ term and ${(\frac{n+1}{2}+1)}^{th}$ or ${\left(\frac{n+3}{2}\right)}^{\text{th }}$ term.

Q.4. What is a coefficient in the Binomial Theorem?
Ans: The coefficients in the binomial expansion are
${}^n{C}_0,{}^n{C}_1,{}^n{C}_2,\dots \dots ,{}^n{C}_r\dots \dots ,{}^n{C}_n$
The coefficients values can be determined by using the formula ${}^n{C}_r=\frac{n!}{r!(n-r)!}$

Q.5. Where is binomial theorem used?
Ans: The binomial theorem is helpful to do the binomial expansion and to find the expansions for the algebraic identities such as
$(a+b{)}^2=a^2+2ab+b^2$
$(a-b{)}^2=a^2-2ab+b^2$
$(a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3$

Learn more about Binomial Theorem

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