These calculus-based techniques evaluate sums involving binomial coefficients multiplied by polynomial weights.

Starting identity: (1+x)^n = sum C(n,r)*x^r.

Differentiation: n(1+x)^{n-1} = sum rC(n,r)x^{r-1}. Substituting x=1: n2^{n-1} = sum rC(n,r). This is the most frequently used identity.

Multiply by x then differentiate: x*(1+x)^n differentiates to give sums with (r+1) or r weights. By differentiating d/dx[x*(1+x)^n] = (1+x)^n + nx(1+x)^{n-1}: this gives sum (r+1)*C(n,r)*x^r. At x=1: (n+2)*2^{n-1}.

Higher derivatives for r^k weights: For sum r^2*C(n,r), use r^2 = r(r-1) + r. Then sum r(r-1)*C(n,r) = n(n-1)*2^{n-2} (from second derivative at x=1). Total: n(n-1)2^{n-2} + n2^{n-1} = n(n+1)*2^{n-2}.

Integration: integral_0^1 (1+x)^n dx = sum C(n,r)/(r+1). LHS = (2^{n+1}-1)/(n+1). This evaluates sums with 1/(r+1) denominators.

Alternating weighted sums: Substitute x=-1 after differentiation. sum (-1)^rrC(n,r) = 0 (from n*(1-1)^{n-1} = 0 for n >= 2).

Combined technique for sum r*C(n,r)*a^r: Differentiate (1+x)^n, multiply by x, then put x=a. Result: na(1+a)^{n-1}.

JEE tip: Whenever you see r, r^2, r(r-1), or 1/(r+1) multiplied by C(n,r), think differentiation or integration. The substitution rC(n,r) = nC(n-1,r-1) is equally powerful and sometimes faster.