Sum Of Binomial Coefficients

Question

The combinatorial sum C(n,0) + C(n,1) + C(n,2) + ... + C(n,n), formed from a complete row of binomial coefficients, evaluates to which closed-form expression?

RankGuru · Accepted Answer

A foundational identity in the binomial theorem states that the sum of all coefficients in a row of Pascal's triangle equals 2^n, derived by a clever choice of variable, which JEE Advanced uses constantly. Start from the expansion (1 + x)^n = C(n,0) + C(n,1)x + C(n,2)x^2 + ... + C(n,n)x^n. Substituting x = 1 makes every power of x equal to one, so the left side becomes (1 + 1)^n = 2^n and the right side becomes exactly the required sum of coefficients. Hence the sum equals 2^n. Option 2^{n-1} represents only the sum of the even-indexed or odd-indexed coefficients, which is half the total. Option n^2 has no basis in the binomial structure. Option n! confuses arrangements with the coefficient sum. The substitution x = 1 is the canonical proof technique here. Plausibility check: for a small case n = 3, the coefficients 1, 3, 3, 1 sum to 8 = 2^3, confirming the formula and the doubling behaviour as n increases by one.

Sum Of Binomial Coefficients

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