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Question

Evaluate:
12c1+22c2+32c3+42c4+......+n2cn

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Solution

s=12C1+22C2+32C3+42C4+.....+n2Cn

(1+x)n=nC0+nC1x+nC2x2+nC3x3+.....+nCnxn

Differentiable w.r.t. x

n(1+x)n1=0+nC1+2nC2x+3nC3x2+.....+nnCxxn1

Multiply whole expression with x

nx(1+x)n1=nC1x+2nC2x2+3nC3x3+...+nnCnxn

Differentiable w.r.t. x again

n(1+x)n1+nx(n1)(1+x)n2=nC1+22nC2x+32nC3x2+nnCnxn1

Put x=1

S=nC1+22nC2+32nC3+....+n2nCn

=n(1+1)n1+n.1.(n1)(1+1)n2

=n2n1+n(n1)2n2

=n2n2(2+(n1))=n2n2(n+1).

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