Greatest Binomial Coefficient

On this page
  1. Direct answer
  2. What you must remember
  3. Greatest coefficient versus greatest term
  4. Three questions that look identical
  5. Frequently asked questions
  6. Related topics

Direct answer

The coefficients of (1 + x)^n climb to a crest and fall back symmetrically: since C(n, r+1)/C(n, r) = (n − r)/(r + 1), they rise while r < (n − 1)/2 and decline after. So the greatest binomial coefficient is C(n, n/2) alone when n is even, and the equal pair C(n, (n−1)/2) = C(n, (n+1)/2) when n is odd — a property of Pascal's row only, with no dependence on x. The numerically greatest term of an expansion, by contrast, depends on |y/x| and is located by a different ratio test; conflating the two is the chapter's most rewarded confusion.

What you must remember

  • Monotonicity: C(n, r+1)/C(n, r) = (n − r)/(r + 1); coefficients increase while r < (n − 1)/2 and decrease afterwards — no full row needs computing.
  • Greatest coefficient: n even → C(n, n/2), e.g. C(10, 5) = 252 for (1 + x)^10; n odd → both C(n, (n−1)/2) and C(n, (n+1)/2), e.g. C(15, 7) = C(15, 8) = 6435.
  • Symmetry: C(n, r) = C(n, n − r); the row reads identically forwards and backwards, which is exactly why odd n produces a tie at the crest.
  • Greatest term (the contrast): in (1 + x)^n with x > 0, T_(r+1) ≥ T_r ⟺ [(n − r + 1)/r]·x ≥ 1; the answer moves with x and can sit far from the middle.
  • Row totals: Σ C(n, r) = 2^n and Σ (−1)^r C(n, r) = 0; the greatest coefficient is the largest summand feeding into 2^n.
  • Values worth carrying: C(10, 5) = 252, C(12, 6) = 924, C(14, 7) = 3432, C(15, 7) = 6435 — these reappear as options constantly.
  • Link to middle term: the middle term of (x + y)^n carries the greatest binomial coefficient whatever x and y are; only its numerical size as a term depends on the ratio.

Greatest coefficient versus greatest term

Find the greatest coefficient in (1 + x)^15, then the numerically greatest term in (1 + 2x)^15. First part: n odd, so the maxima are C(15, 7) and C(15, 8), each 6435 — positions 8 and 9 in the expansion. Second part: T(r+1) = C(15, r)·2^r, and the ratio test gives T(r+1)/T_r = [(16 − r)/r]·2 ≥ 1 ⟺ 32 − 2r ≥ r ⟺ r ≤ 32/3. So terms grow up to r = 10 and shrink after: the maximum sits at T₁₁ = C(15, 10)·2^10 = 3003 × 1024 = 3075072, the eleventh term — nowhere near the middle. The coefficient question read only Pascal's row; the term question multiplied the row by 2^r, and that exponential weighting dragged the peak to the right.

The lesson generalises: for (1 + kx)^n the greatest term drifts towards the end as k grows, while the greatest coefficient never moves. Whenever an option set offers "the middle term is the greatest", read carefully whether coefficient or term is meant — both statements can be true or false depending on that word.

Three questions that look identical

Greatest coefficient (parity of n alone), greatest term (depends on |y/x| through the ratio test), and middle term (position by parity) form Advanced's favourite triple — all three can appear in one multiple-correct item. Main dresses them up with substitutions: asking about (2 + x)^15 quietly re-indexes to (1 + x/2)^15 for coefficient questions, where the answer is still parity arithmetic. Traps: reporting one maximum where two exist (odd n always gives a pair); applying the coefficient rule to (2 + 3x)^n as if the row of Pascal still decided it — with constants attached, the "coefficients" include powers of 2 and 3 and the greatest-term logic returns; and ratio slips, using (n − r)/(r + 1) with an x multiplied in when the question was purely about coefficients.

Frequently asked questions

What is the greatest coefficient in (1 + x)^15?

Both C(15, 7) and C(15, 8), each equal to 6435 — odd n always yields two equal maxima at the middle.

Does the greatest coefficient depend on x?

No — it is a property of Pascal's row for n alone; the numerically greatest term is the quantity that depends on x.

How do I locate the numerically greatest term in (1 + x)^n?

Solve [(n − r + 1)/r]·|x| ≥ 1 for the largest admissible r; the maximum term is then T_(r+1).

Why are the two middle coefficients equal for odd n?

Symmetry C(n, r) = C(n, n − r) maps (n − 1)/2 onto (n + 1)/2, so the twin crests match exactly.

Is the middle term of (x + y)^n the one with the greatest coefficient?

Yes — the middle term always carries the greatest binomial coefficient, though not necessarily the numerically greatest term when |y/x| ≠ 1.

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