JEE Main Maths Binomial Theorem 2027: General Term, Middle Term & Coefficients

One formula does nearly everything in the Binomial Theorem chapter: the general term. Find the term you want, match the power of $x$, solve for $r$, and you are done. That makes this chapter fast to learn and quick to score in.

Write the general term first, then hunt for the value of $r$ that gives what the question asks.

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Chapter at a Glance

Priority reflects past-paper trends, not an official NTA weightage.

What the NTA Syllabus Covers

• Binomial theorem for a positive integral index
• General term and middle term
• Simple applications

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Master These Topics

1. The Theorem and the General Term

For a positive integer $n$:

$$(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$$

The general term, the $(r+1)$th term, is

$$T_{r+1} = \binom{n}{r} a^{n-r} b^{r}$$

The expansion has $n + 1$ terms, and the coefficients come straight from Pascal's triangle.

Worked example (a specific coefficient): The coefficient of $x^3$ in $(2 - x)^5$ is $\binom{5}{3} 2^{2} (-1)^3 = 10 \cdot 4 \cdot (-1) = -40$.

Trap: The $(r+1)$th term uses index $r$, not $r + 1$. Mixing this up shifts every answer by one term.

2. Term Independent of $x$

Write the general term, collect the powers of $x$ and set the exponent to zero.

Worked example: In $\left(x^2 + \dfrac{1}{x}\right)^9$, the general term is $\binom{9}{r} x^{2(9-r)} x^{-r} = \binom{9}{r} x^{18 - 3r}$. Setting $18 - 3r = 0$ gives $r = 6$, so the term independent of $x$ is $\binom{9}{6} = 84$.

3. Middle Term

• If $n$ is even, there is one middle term, the $\left(\frac{n}{2} + 1\right)$th term.
• If $n$ is odd, there are two middle terms, the $\frac{n+1}{2}$th and $\frac{n+3}{2}$th terms.

Worked example: For $\left(x + \dfrac{1}{x}\right)^{10}$ the middle term is the 6th term, $\binom{10}{5} = 252$, and it is independent of $x$.

4. Coefficient Properties and Approximations

• $\displaystyle\sum_{r=0}^{n}\binom{n}{r} = 2^n$ and $\binom{n}{r} = \binom{n}{n-r}$
• Pascal's rule: $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$
• The sum of all coefficients of a polynomial expansion is found by putting $x = 1$. For $(1 + 2x)^5$ it is $3^5 = 243$.

For small $x$, $(1 + x)^n \approx 1 + nx + \dbinom{n}{2}x^2$.

Worked example: $(1.01)^{10} = (1 + 0.01)^{10} \approx 1 + 0.1 + 45(0.0001) = 1.1045$, and the exact value is about $1.1046$.

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Common Traps to Avoid

• Using $r$ as the term number. The $(r+1)$th term has index $r$.
• Forgetting to include the sign when $b$ is negative.
• Missing that odd $n$ has two middle terms.
• Not setting the exponent of $x$ to zero when asked for the constant term.

60-Second Revision Sheet

• $T_{r+1} = \binom{n}{r} a^{n-r} b^r$
• Middle term: $n$ even gives the $(\tfrac{n}{2}+1)$th term; $n$ odd gives two terms
• $\sum \binom{n}{r} = 2^n$; sum of coefficients: put $x = 1$
• $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$

Your Study Plan

1. Day 1: the theorem, general term and simple coefficient problems.
2. Day 2: term independent of $x$ and terms with fractional powers.
3. Day 3: middle terms, sums of coefficients and binomial identities.
4. Day 4: approximations and a timed mixed set.

Practice Binomial Theorem Questions Free → (opens in a new tab)

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Continue Your Maths Journey

• Previous chapter: Permutations & Combinations
• Next chapter: Sequence & Series
• All 14 JEE Main Maths chapters
• Complete JEE Main Syllabus 2027 guide

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