# Middle Term of a Binomial Expansion

> Middle term of a binomial expansion in JEE Mathematics: general term, middle-term rules for even and odd n, terms from the end and independent-term finds.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/binomial-middle-term
- 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: "Middle Term of a Binomial Expansion", PrepElephant, https://prepelephant.com/topics/jee/mathematics/binomial-middle-term

## Direct answer

Whether (x + y)^n has one middle term or two is a parity question answered before any expansion: n even gives the single middle term T_(n/2 + 1), while n odd gives the pair T_((n+1)/2) and T_((n+3)/2) with equal coefficients. The engine behind every such question is the general term T_(r+1) = C(n, r) x^(n−r) y^r, which also locates the term independent of x (set the net power of x to zero), the coefficient of x^m (solve n − r = m), and the kth term from the end — no full expansion ever needed. Since (x + y)^n has exactly n + 1 terms, position arithmetic is always exact.

## What you must remember

- **General term:** T_(r+1) = C(n, r) x^(n−r) y^r for (x + y)^n; for (x − y)^n the sign rides on y, giving T_(r+1) = (−1)^r C(n, r) x^(n−r) y^r.
- **Middle term:** n even → T_(n/2 + 1) (so (x + y)^10 peaks at T₆); n odd → T_((n+1)/2) and T_((n+3)/2) (so (x + y)^11 peaks at T₆ and T₇), the twin coefficients equal by C(n, r) = C(n, n − r).
- **Term from the end:** the kth term from the end of (x + y)^n equals the kth term from the beginning of (y + x)^n, i.e. the (n − k + 2)th term from the start.
- **Term independent of x:** in expansions like (x^p + x^(−q))^n, set the net power p(n − r) − qr = 0 in T_(r+1) and solve for r; a non-integer r means no such term exists.
- **Coefficient of x^m:** solve n − r = m, so r = n − m and the coefficient is C(n, n − m) — one substitution, no expansion.
- **Count check:** (x + y)^n has n + 1 terms and the exponents in every term sum to n — a fast sanity test on any claimed term.
- **Numerically greatest term:** compare neighbours via T_(r+1)/T_r = [(n − r + 1)/r]·|y/x| ≥ 1 — a different question from the middle term, since it depends on |y/x|.

## Hunting a specific term

Find the term independent of x in (x² + 1/(2x))⁹. The general term is T_(r+1) = C(9, r)(x²)^(9−r)·(2x)^(−r) = C(9, r)·2^(−r)·x^(18 − 3r). Independence demands 18 − 3r = 0, so r = 6 — a whole number, which certifies the term exists. The term is T₇ = C(9, 6)/2⁶ = 84/64 = 21/16. Notice the two-part discipline: first the power equation (which decides existence), then the coefficient arithmetic — reversing the order wastes time when r fails to come out whole.

The same template handles the middle term. For (x + y)^10, n is even, so the middle term is T₆ = C(10, 5)x⁵y⁵ = 252x⁵y⁵. For (x − y)^11, n is odd: T₆ and T₇ are the middle terms, and the (−1)^r sign makes T₆ = −C(11, 5)x⁶y⁵ while T₇ = +C(11, 6)x⁵y⁶ — the signs alternate around the middle even though the magnitudes match in pairs by symmetry.

## How the exam frames it

Main asks the middle term or the independent term directly — single-correct, two lines of work, answer options built from sign and parity slips. Advanced prefers multiple-correct grids: "which of the following expansions contain a term free of x?", where each option is a different (p, q, n) triple and only the power equation p(n − r) = qr decides membership. The standing traps: forgetting (−1)^r in (x − y)^n, so a negative coefficient option gets marked wrong; and answering a coefficient where a term was asked, or vice versa — the paper sets the two nouns side by side deliberately. A third habit worth building: always state the term number T_(r+1), not just r, because "which term" questions grade the index.

## Frequently asked questions

### What is the middle term of (x + y)^n for odd n?

Two middle terms, T_((n+1)/2) and T_((n+3)/2), with equal coefficients because C(n, (n−1)/2) = C(n, (n+1)/2) by symmetry.

### How do I find the term independent of x in (x^p + x^(−q))^n?

Set the net power p(n − r) − qr = 0 in T_(r+1) and solve for r; the term exists only if r is a whole number within 0 to n.

### Is the middle term always the numerically greatest term?

No — the greatest term also depends on |y/x| through the ratio test, while the middle term's position depends only on n.

### What is the kth term from the end of an expansion?

The kth term from the end of (x + y)^n is the kth term of (y + x)^n read forwards, i.e. the (n − k + 2)th from the beginning.

### Does (x − y)^n change the middle term position?

No — parity fixes the position identically; only the sign (−1)^r changes, making every odd-r term negative.