# Binomial Theorem

> Binomial theorem for JEE Mathematics; general term, middle term, term independent of x and coefficient sums with exam-focused practice tips.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/binomial-theorem
- Exam / course: JEE · Subject: Mathematics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Binomial Theorem", PrepElephant, https://prepelephant.com/topics/jee/mathematics/binomial-theorem

## Direct answer

For a positive integer n, (x + y)^n expands as the sum of C(n, k) x^(n - k) y^k for k running from 0 to n, giving n + 1 terms. The single most used line in JEE is the general term T(k + 1) = C(n, k) x^(n - k) y^k: write the powers of x and y in terms of k, then impose the condition asked — a specific power, the middle term or a term independent of x.

## What you must remember

- Expansion: (x + y)^n = sum over k from 0 to n of C(n, k) x^(n - k) y^k; (1 + x)^n begins 1 + nx + C(n, 2) x^2 + ...
- General term: T(k + 1) = C(n, k) x^(n - k) y^k; the k-shift (the rth term uses k = r - 1) is where most marks are lost.
- Middle term: for even n a single middle term with k = n/2; for odd n two middle terms, with k = (n - 1)/2 and k = (n + 1)/2.
- Term independent of x: set the net power of x in T(k + 1) to zero and solve for k; k must come out a non-negative integer.
- Coefficient sums: putting x = y = 1 gives the total 2^n; putting x = 1, y = -1 splits even and odd position sums, each 2^(n - 1).
- Weighted sums: the sum of k C(n, k) over k equals n × 2^(n - 1); the greatest binomial coefficient sits at the middle, k = n/2.
- Approximation and divisibility: (1 + x)^n is close to 1 + nx for small x; for divisibility, write the base as a sum near a multiple of the divisor and expand — every term except the last then divides.

## Common confusion

The perennial slip is the index shift: the term in x^m is not the mth term. Set the power of x to n - k = m, get k, then call it the (k + 1)th term. Also distinguish the greatest coefficient, which depends only on n, from the numerically greatest term, which depends on x as well — for the latter, compare successive term ratios with 1 rather than quoting the middle term.

## Exam-focused takeaway

JEE Main asks the general term for a specific power, the term independent of x, middle terms and coefficient sums — fast numericals. JEE Advanced prefers coefficient extraction in products of expansions, identities built on the weighted sums, and divisibility arguments from the expansion. The routine is identical: write T(k + 1) in full, turn the condition into an equation in k, check integrality.

## Frequently asked questions

### How do I find the term independent of x?

Set the total power of x in T(k + 1) to zero and solve for k; a non-negative integer k gives the required term.

### Which term is the middle term of (x + y)^10?

With n = 10 even, the single middle term is the 6th, at k = 5.

### What is the sum of all binomial coefficients in an expansion?

2^n, by substituting x = y = 1; the even and odd position sums are each 2^(n - 1).

### How do I find the coefficient of x^m in a product of expansions?

Write the general term of each factor and match powers so they sum to m — a small system.

### What is the difference between the greatest coefficient and the greatest term?

The greatest coefficient depends only on n and sits at the middle; the numerically greatest term depends on x too — compare consecutive term ratios with 1.