Binomial Coefficient Identities
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Direct answer
Adding the n + 1 entries of the nth row of Pascal's triangle gives 2^n — the first of the summation identities the exam treats as vocabulary. The core set: Σ r·nCr = n·2^(n−1); Σ r²·nCr = n(n + 1)·2^(n−2); the alternating row sums to zero, so even-indexed and odd-indexed sums are each 2^(n−1); Σ (nCr)² = 2nCn; and the hockey stick Σ (r from k to n) rCk = (n + 1)C(k + 1). Every one is derived, not memorised, from a single machine operating on (1 + x)^n: substitute x = 1, x = −1, differentiate, multiply by x and differentiate again, or integrate — which is why the identities survive exam pressure even when memory does not.
What you must remember
- Row sum: Σ (r = 0 to n) nCr = 2^n from x = 1; alternating sum Σ (−1)^r nCr = 0 from x = −1, splitting the row into two equal halves of 2^(n−1).
- First moment: Σ r·nCr = n·2^(n−1) — differentiate (1 + x)^n, multiply by x, set x = 1.
- Second moment: Σ r²·nCr = n(n + 1)·2^(n−2); the third follows the same pattern, Σ r³·nCr = n²(n + 3)·2^(n−3).
- Sum of squares: Σ (nCr)² = 2nCn — choosing r from n twice, or the coefficient of x^n in (1 + x)^n(1 + x)^n.
- Hockey stick: Σ (r = k to n) rCk = (n + 1)C(k + 1) — summing down a diagonal of Pascal's triangle collapses to one entry.
- Integrated identity: Σ nCr/(r + 1) = (2^(n + 1) − 1)/(n + 1), from integrating (1 + x)^n over [0, 1].
- Vandermonde in general: Σ rCk × (n − r)C(m − k) = nCm, the committee-counting identity behind most combinatorial proofs.
One identity, two derivations
Evaluate Σ r(n − r)·nCr. First route, by decomposition: r·nCr = n·(n − 1)C(r − 1), so the sum becomes n Σ (n − r)(n − 1)C(r − 1); shifting the index with s = r − 1 leaves (n − 1 − s) beside (n − 1)C(s), and the two standard sums give n[(n − 1)·2^(n − 1) − (n − 1)·2^(n − 2)] = n(n − 1)·2^(n − 2). Second route, by symmetry: r(n − r) pairs each entry with its mirror, and the answer must be half of Σ [r² + (n − r)² − (r − (n − r))²]-style bookkeeping — messier, which is itself the lesson: the differentiation machine beats cleverness. Verify at n = 3: direct computation gives 6 + 6 = 12, and n(n − 1)2^(n − 2) = 3 × 2 × 2 = 12. The numerical spot-check at a tiny n is the professional habit — thirty seconds that catch every misremembered exponent before they cost a mark.
Derive, do not memorise
JEE Main asks the plug-in identities: given n = 10, report Σ r·10Cr = 10 × 2⁹ = 5120 — where the distractors are n·2^n (10240) and (n − 1)·2^(n−1) (2304), one exponent slip away in each direction. The alternating identities appear as "sum of coefficients of even powers", answer 2^(n−1), with 2^n planted beside it. JEE Advanced asks for derivations under pressure: mixed sums like Σ r(n − r) nCr above, or (1 + x)^n expansions where a coefficient comparison replaces algebra — the coefficient of x^n in (1 + x)^n(1 + x)^n being the sum-of-squares route. The hockey stick shows up as Σ (r = 2 to 20) rC2 = 21C3 = 1330, computable only if the identity is recognised; the trap is off-by-one in the upper index, with 22C3 = 1540 sitting in the options beside it. The meta-skill this chapter teaches: locate each identity as one operation on the generating function — x = 1 for plain sums, differentiation for r-weights, x·(d/dx) twice for r², integration for the (r + 1) denominator — and the entire list compresses to four moves.
Frequently asked questions
What is the sum of all binomial coefficients in row n?
2^n, from setting x = 1 in (1 + x)^n; the alternating sum with x = −1 is zero.
What is Σ r·nCr?
n·2^(n−1), obtained by differentiating (1 + x)^n, multiplying by x and setting x = 1.
What is the sum of squares of the coefficients in row n?
2nCn — equivalently the coefficient of x^n in (1 + x)^n(1 + x)^n.
What does the hockey stick identity state?
Σ (r = k to n) rCk = (n + 1)C(k + 1): a diagonal of Pascal's triangle sums to the entry just below its last term.
How is Σ nCr/(r + 1) computed?
Integrate (1 + x)^n from 0 to 1: the result is (2^(n + 1) − 1)/(n + 1).