Integration Application

Series in which binomial coefficients are divided by successive integers are evaluated by integrating the expansion term by term, a technique that complements differentiation in JEE Advanced. Consider (1 - x)^n = sum_{r=0}^{n} C(n, r)(-1)^r x^r and integrate both sides from 0 to 1. The left integral is integral_0^1 (1 - x)^n dx = [-(1 - x)^{n+1}/(n+1)]0^1 = 1/(n+1). The right side integrates term by term to sum{r=0}^{n} C(n, r)(-1)^r /(r+1), which is exactly C(n,0)/1 - C(n,1)/2 + C(n,2)/3 - ... Hence the series equals 1/(n+1). Option 1/n drops the plus one that arises from raising the power during integration. Option 1/(n-1) mishandles the limits. Option 2^n/(n+1) wrongly multiplies by a coefficient sum. The integration limits 0 to 1 are chosen precisely to produce the reciprocal integer denominators, because integrating x^r over this interval yields exactly 1/(r+1). The legitimacy of integrating the finite sum term by term is guaranteed since a polynomial is a finite combination of continuous functions, so no convergence subtlety arises. Plausibility check: for n = 1 the series is 1 - 1/2 = 1/2 = 1/(1 + 1), and for n = 2 it is 1 - 1 + 1/3 = 1/3 = 1/(2 + 1), confirming the formula across small cases.