

Introduction to Middle -Term Formula for a Binomial Expansion:
The middle-term formula is a formula used to find the middle term of a binomial expansion. Recall that a binomial expansion is an expression involving the sum or difference of two terms raised to some integral power. Depending on the total number of terms, we can write the middle term of that expression.
Definition of Binomial Expansion
A binomial expansion is an expansion of the sum or difference of two terms raised to some power n, where n is a positive integer. The general form of binomial expansion is \[{(a + b)^n}\], where a and b are two algebraic terms with different variables. They may be numbers as well.
Some examples of Binomial Expansion are: \[{(2x + 3y)^5},{(5{x^2}y - 4x{y^2})^3},{(x + {\textstyle{1 \over {2x}}})^{10}}\]and so on.
Binomial Theorem for a Positive Integer n
The Binomial Theorem gives the expansion of a binomial to any positive integral power. It is stated as:
\[{(a + b)^n} = {}^n{C_0}{a^{n - 0}}{b^0} + {}^n{C_1}{a^{n - 1}}{b^1} + {}^n{C_2}{a^{n - 2}}{b^2} + \cdots + {}^n{C_r}{a^{n - r}}{b^r} + \cdots + {}^n{C_n}{a^{n - n}}{b^n}\]. This means that
There are (n+1) terms in a binomial expansion.
The compact form of binomial theorem is: \[{(a + b)^n} = \sum\limits_{k = 0}^n {{}^n{C_k}{a^{n - k}}{b^k}} \] .
The index of the first term, i.e., ‘a’ decreases whereas the index of the second term, i.e., ‘b’ increases as we go from left to right in binomial expansion.
The coefficients of the terms\[{}^n{C_r}\]are known as Binomial Coefficients.
For Example, the binomial expansion of \[{(2x + 3y)^5}\] is \[\begin{array}{l}{(2x + 3y)^5} = {}^5{C_0}{(2x)^{5 - 0}}{(3y)^0} + {}^5{C_1}{(2x)^{5 - 1}}{b^1} + {}^5{C_2}{(2x)^{5 - 2}}{b^2}\\ + {}^5{C_3}{(2x)^{5 - 3}}{(3y)^3} + {}^5{C_4}{(2x)^{5 - 4}}{(3y)^4} + {}^5{C_5}{(2x)^{5 - 5}}{(3y)^5}\\ = {(2x)^5} + 5{(2x)^4}{(3y)^1} + 10{(2x)^3}{(3y)^2} + 10{(2x)^2}{(3y)^3} + 5{(2x)^1}{(3y)^4} + {(3y)^5}\\ = 32{x^5} + 240{x^4}y + 720{x^3}{y^2} + 1080{x^2}{y^3} + 810x{y^4} + 243{y^5}\end{array}\]
General Term of Binomial Expansion
The general rth term of a Binomial Expansion \[(a+b)^n\] is denoted by \[{T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}\].
For Example, Given Binomial Expansion \[{(2x + 3y)^5}\], the general term is \[{T_{r + 1}} = {}^5{C_r}{(2x)^{5 - r}}{(3y)^r}\]. If we need to find the third term, equate 3 to r+1, which means
\[r + 1 = 3 \Rightarrow r = 2\]. So,
\[\begin{array}{l}{T_{r + 1}} = {}^5{C_2}{(2x)^{5 - 2}}{(3y)^2} = \dfrac{{5!}}{{2!\left( {5 - 2} \right)!}}{(2x)^3}{(3y)^2}\\ = \dfrac{{5!}}{{2!\left( 3 \right)!}}{(2x)^3}{(3y)^2}\\ = 10{(2x)^3}{(3y)^2}\\ = 720{x^3}{y^2}\end{array}\]
Middle Term of Binomial Expansion
The Middle Term of a Binomial Expansion is the term that comes in the middle of the expansion of the binomial, given by the Binomial Theorem. If the power of expansion of the sum or difference of two terms is an even number, then there is only one middle term. Otherwise, there are two middle terms.
How To Find Middle Term?
The middle term of a Binomial Expansion depends on the total number of terms in the given expansion. If n is even, (n+1) is odd so that the expansion will have an odd number of terms. In this case, the middle term is \[{(\dfrac{n}{2} + 1)^{th}}\] term. If n is odd, (n+1) will be even. So, there will be two middle terms, i.e., \[{(\dfrac{{n + 1}}{2})^{th}}\] term and \[{(\dfrac{{n + 1}}{2} + 1)^{th}}\]term. We will use the general term formula and put the value of r as \[{\textstyle{n \over 2}}\], if n is even to find the middle term. This is illustrated by the following diagram:

Middle Terms of a Binomial Expansion
Middle Term Example
Let us understand this by an example. Given a Binomial Expansion \[{(x + {\textstyle{1 \over x}})^4}\], n=4 which is even. So, n+1=4+1=5 is odd. So, the middle term is \[{(\dfrac{4}{2} + 1)^{th}} = {(2 + 1)^{th}} = {3^{rd}}\]term, which is
\[\begin{array}{l}{T_3} = {T_{2 + 1}} = {}^4{C_2}{(x)^{4 - 2}}{({\textstyle{1 \over x}})^2}\\ = \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}}{x^2}.\dfrac{1}{{{x^2}}}\\ = 6\end{array}\]
For Binomial Expansion \[{(2x - y)^7}\], n=7, so, n+1=8 which is even. The middle terms are \[{(\dfrac{{7 + 1}}{2})^{th}} = {(\dfrac{8}{2})^{th}} = {4^{th}}\]and \[{(\dfrac{{7 + 1}}{2} + 1)^{th}} = {(\dfrac{8}{2} + 1)^{th}} = {(4 + 1)^{th}} = {5^{th}}\], i.e., \[\begin{array}{l}{T_4} = {T_{3 + 1}} = {}^7{C_3}{(2x)^{7 - 3}}{( - y)^3}\\ = \dfrac{{7!}}{{3!\left( {7 - 3} \right)!}}(2x)^4({-y^3})\\ =-560{x^4}{y^3}\end{array}\]
and \[\begin{array}{l}{T_5} = {T_{4 + 1}} = {}^7{C_4}{(2x)^{7 - 4}}{( - y)^4}\\ = \dfrac{{7!}}{{4!\left( {7 - 4} \right)!}}{(2x)^3}{y^4}\\ = 35(8{x^3}){y^4}\\ = 280{x^3}{y^4}\end{array}\]
Interesting Facts
The Binomial Theorem is used in Probability and Statistics. Our economy depends on it.
It is used for finding higher power roots of equations.
It is used to predict the weather and for giving rank to candidates.
Solved Examples
1. Find the middle terms in the expansion of (3-x3)7.
Ans: Given Binomial Expansion is (3-x3)7.
Here, n=7. So, n+1=8, which is even. The middle terms are \[{(\dfrac{{7 + 1}}{2})^{th}} = {(\dfrac{8}{2})^{th}} = {4^{th}}\]and \[{(\dfrac{{7 + 1}}{2} + 1)^{th}} = {(\dfrac{8}{2} + 1)^{th}} = {(4 + 1)^{th}} = {5^{th}}\], i.e.,
\[\begin{array}{l}{T_4} = {T_{3 + 1}} = {}^7{C_3}{(3)^{7 - 3}}{( - {x^3})^3}\\ = \dfrac{{7!}}{{3!\left( {7 - 3} \right)!}}(3^4)( - {x^9})\\ = - 2835{x^9}\end{array}\]
and
\[\begin{array}{l}{T_5} = {T_{4 + 1}} = {}^7{C_4}{(3)^{7 - 4}}{( - {x^3})^4}\\ = \dfrac{{7!}}{{4!\left( {7 - 4} \right)!}}{(3)^3}{({x^3})^4}{( - 1)^4}\\ = 35(27)({x^{12}})(1)\\ = 945{x^{12}}\end{array}\]
The middle term are \[945{x^{12}}\]
2. Show that the Middle Term in the expansion of (1+x)2m is \[\dfrac{{1.3.5.7...(2m - 1)}}{{m!}}{2^m}{x^m}\] where m is a positive integer.
Ans: Given Binomial Expansion is (1+x)2m.
Step 1:
Finding the number of terms of the middle term:
Here, n=2m. So, n+1=2m+1, which is odd. So, the middle term is \[{(\dfrac{{2m}}{2} + 1)^{th}} = {(m + 1)^{th}}\]term.
The general term of \[(r+1)^{th}\] of \[(1+x)^n\] is \[{T_{r + 1}} = {}^n{C_r}{x^r}\]
Step 2:
Compute \[{(m + 1)^{th}}\]term of given expansion using the formula of general term:
\[{T_{m + 1}} = {}^{2m}{C_m}{(1)^{2m - m}}{(x)^m}\]
\[\begin{array}{l} = {}^{2m}{C_m}{(1)^{2m - m}}{x^m}\\ = \dfrac{{(2m)!}}{{m!\left( {2m - m} \right)!}}{(1)^m}{x^m}\end{array}\]
\[\begin{array}{l} = \dfrac{{(2m)(2m - 1)(2m - 2)...(m + 1)m(m - 1)...2.1}}{{m!m!}}(1){x^m}\\ = \dfrac{{(2m)(2m - 1)(2m - 2)...2.1}}{{m!m!}}{x^m}\end{array}\]
\[\begin{array}{l} = \dfrac{{\{ 2m(2m - 2)(2m - 4)...(2m - (2m - 2))\} \{ (2m - 1)(2m - 3)....(3)(1)\} }}{{m!m!}}{x^m}\\ = \dfrac{{\{ 1.3.5...(2m - 1)\} \{ m(m - 1)(m - 2)...3.2.1\} {2^m}{x^m}}}{{m!m!}}\\ = \dfrac{{m!\{ 1.3.5...(2m - 1)\} {2^m}{x^m}}}{{m!m!}}\\ = \dfrac{{1.3.5...(2m - 1){2^m}{x^m}}}{{m!}}\end{array}\]
3. Find the value of α for which the coefficients of middle terms in the expansion of (1+αx)4 and (1-αx)6 are equal.
Ans: Consider the expression (1+αx)4. Here, n=4. So, n+1=4+1=5, and 5 is odd. The middle term is \[{(\dfrac{4}{2} + 1)^{th}} = {(2 + 1)^{th}} = {3^{rd}}\]term.
Now, T3=T2+1
\[\begin{array}{l} = {}^4{C_2}{(1)^{4 - 2}}{(\alpha x)^2}\\ = \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}}{(1)^2}{\alpha ^2}{x^2}\\ = 6{\alpha ^2}{x^2}\end{array}\]
Now, consider the expression (1-αx)6. Here, n=6. So, n+1=6+1=7, and 7 is odd. The middle term is \[{(\dfrac{6}{2} + 1)^{th}} = {(3 + 1)^{th}} = {4^{th}}\]term.
T4=T3+1
\[\begin{array}{l} = {}^6{C_3}{(1)^{6 - 3}}{( - \alpha x)^3}\\ = \dfrac{{6!}}{{3!\left( {6 - 3} \right)!}}{(1)^3}( - {\alpha ^3}{x^3})\\ = - 20{\alpha ^3}{x^3}\end{array}\]
Since the coefficients of both the middle terms are equal, so comparing both, we get
6α2 = -20α3
\[\begin{array}{l} \Rightarrow - \dfrac{6}{{20}} = \dfrac{{{\alpha ^3}}}{{{\alpha ^2}}}\\ \Rightarrow - \dfrac{3}{{10}} = \alpha \\ \Rightarrow \alpha = - \dfrac{3}{{10}}\end{array}\]
The value of α for which the coefficients of both the middle terms are equal is \[( - \dfrac{3}{{10}})\].
Key Features
The position of the middle term of a binomial expansion depends on the total number of terms contained in the expansion(n+1).
If n is odd, the Binomial Expansion will have two middle terms: \[{(\dfrac{{n + 1}}{2})^{th}}\] and \[{(\dfrac{{n + 1}}{2} + 1)^{th}}\].
If n is even, the Binomial Expansion will have one middle term: \[{(\dfrac{n}{2} + 1)^{th}}\] .
Practice Questions
1. Find the middle terms in the expansion of
\[\left ( \dfrac{2x}{3} - \dfrac{3}{2x} \right )^{20}\]
\[\mathrm {(1-2x+x^2)}^{10}\]
2. Prove that the coefficient of the middle term in the expansion of \[{(1+x)}^{2n}\] is equal to the sum of coefficients of the middle terms of the expansion \[{(1+x)}^{2n-1}\].
FAQs on Middle Term Formula
1. What are the formulas to find the middle term in the binomial expansion of (a + b)ⁿ?
The formula to find the middle term in a binomial expansion depends on whether the power 'n' is even or odd. The rules are as follows:
If 'n' is an even number, there is only one middle term, which is the (n/2 + 1)th term of the expansion.
If 'n' is an odd number, there are two middle terms: the ((n+1)/2)th term and the ((n+1)/2 + 1)th term.
2. How do you find the middle term for an expansion like (2x - y)⁶?
In this case, the power n=6, which is an even number. The total number of terms in the expansion will be n+1 = 7. Since there is an odd number of terms, there will be a single middle term.
The position of the middle term is calculated using the formula (n/2 + 1), which is (6/2 + 1) = 4. So, we need to find the 4th term.
Using the general term formula Tr+1 = nCr an-r br, the 4th term (T3+1) is: 6C3 (2x)6-3 (-y)3.
This simplifies to 20 × (8x³) × (-y³) = -160x³y³.
3. Why does the binomial expansion of (a + b)ⁿ have two middle terms when 'n' is odd?
The total number of terms in any binomial expansion is always n + 1. When the power 'n' is an odd number, the total number of terms (n+1) becomes an even number. Any sequence with an even number of items does not have a single central value; instead, it has a pair of terms in the middle. For instance, if n=5 (odd), the expansion has 5+1=6 terms, and both the 3rd and 4th terms are considered middle terms.
4. How is finding the middle term in a binomial expansion different from finding it in an Arithmetic Progression (AP)?
Although both concepts involve a "middle term," they are fundamentally different:
In a Binomial Expansion, the middle term is a specific algebraic expression determined by the exponent 'n'. Its calculation involves combinatorial formulas (nCr) and depends on the terms within the binomial (a and b).
In an Arithmetic Progression (AP), the middle term is a specific number in a sequence. Its value is determined by its position and can be found using the AP formula a + (k-1)d or by averaging the first and last terms if there's an odd number of terms in the sequence.
5. Can you use the middle term formula if the power 'n' in (a+b)ⁿ is negative or a fraction?
No, the standard middle term formulas are only applicable when 'n' is a positive integer. This is because a positive integer 'n' results in a finite expansion with (n+1) terms, which has a defined middle. When 'n' is negative or a fraction, the binomial theorem produces an infinite series, which does not have a "middle term" in the same way a finite series does.
6. What is the significance of the middle term in the expansion of an expression like (x + 1/x)2n?
In specific algebraic expressions, the middle term holds special importance. For the expansion of (x + 1/x)2n, the middle term is significant because it is the term independent of the variable x. This happens because the powers of 'x' and '1/x' cancel each other out perfectly in that specific term, making its value a constant. This property is often used to solve specific types of problems in algebra.
7. Is the coefficient of the middle term always the largest in a binomial expansion?
Yes, for the expansion of (a+b)ⁿ, the binomial coefficient (the nCr part) of the middle term(s) is the largest. When 'n' is even, there is one middle term, and its coefficient, nCn/2, is the greatest. When 'n' is odd, there are two middle terms with equal coefficients, nC(n-1)/2 and nC(n+1)/2, and these are the greatest coefficients in the expansion.





