Middle Term of a Binomial Expansion
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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.