Weighted Coefficient Sum

Sums where each binomial coefficient is weighted by its index are evaluated by differentiating the binomial expansion and then substituting a value, a calculus-flavoured JEE Advanced identity. Start from (1 + x)^n = sum_{r=0}^{n} C(n, r) x^r and differentiate both sides with respect to x to obtain n(1 + x)^{n-1} = sum_{r=1}^{n} r C(n, r) x^{r-1}. Now substitute x = 1, which gives n(1 + 1)^{n-1} = n 2^{n-1} on the left, while the right side becomes sum_{r=1}^{n} r C(n, r) = C(n,1) + 2C(n,2) + ... + n C(n,n). Therefore the weighted sum equals n 2^{n-1}. Option 2^n is the unweighted coefficient sum and ignores the index multipliers. Option n 2^n overstates the power of two by failing to reduce the exponent after differentiation. Option (n-1)2^{n-1} miscounts the leading factor. Differentiation is the essential step that injects the index r into each term, since the derivative of x^r is r x^{r-1} and it is precisely this factor r that converts the plain coefficients into index-weighted ones. The same identity can be read combinatorially as choosing a committee of any size from n people and then designating one chairperson, which counts to n 2^{n-1} as well. Plausibility check: for n = 3 the sum 1·3 + 2·3 + 3·1 = 12 = 3·2^{2} = n 2^{n-1}, confirming the formula.