Alternating Sum Of Coefficients

The alternating sum of binomial coefficients is evaluated by substituting a negative value into the standard expansion, a method JEE Advanced expects students to recognise instantly. Begin with (1 + x)^n = C(n,0) + C(n,1)x + C(n,2)x^2 + ... + C(n,n)x^n and put x = -1. The left side becomes (1 - 1)^n = 0^n, which equals 0 for every positive integer n. The right side becomes C(n,0) - C(n,1) + C(n,2) - C(n,3) + ..., exactly the alternating sum in question. Therefore the alternating sum is 0. Option 1 ignores that the base 1 - 1 vanishes. Option 2^n is the result of substituting x = +1, not x = -1, so it answers a different question. Option (-1)^n incorrectly treats the base as -1 raised to n rather than (1 - 1)^n. This zero result is equivalent to saying the number of even-sized subsets equals the number of odd-sized subsets. Plausibility check: for n = 2 the coefficients give 1 - 2 + 1 = 0, and for n = 3 they give 1 - 3 + 3 - 1 = 0, confirming the cancellation in general.