Q.
The expression nCo+2nC1+3nC2+.....+(n+1)nCn is equal to
2710
195
AMUAMU 2011Permutations and Combinations
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Solution:
We know, (1+x)n=nC0+nC1x+nC2x2 +...+nCnxn ⇒x(1+x)n=nC0x+nC1x2+nC2x3+... +nCnxn+1
On differentiating w.r.t. x, we get (1+x)n+nx(1+x)n−1=nC0+2⋅nC1x +3⋅nC2x2+...+(n+1)⋅nCnxn
Put x=1, we get 2n+2n−1⋅n=nC1+2⋅nC1+3⋅nC2 +...+(n+1)⋅nCn =(n+2)2n−1