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Questions  

If nN, n>1, then value of E=anC1(a1)+nC2(a2)++(1)n(an) nCn is

a
a
b
0
c
a2
d
2n

detailed solution

Correct option is B

We can write E asa nC0−nC1+nC2−…+(−1)n nCn+ nC1− (2)  nC2+(3) nC3−…−(−1)n(n) nCn    =0+F whereF= nC1−(2) nC2+(3) nC3−…−  (−1)n(n) nCn       We have (1+x)n=nC0+nC1x+nC2x2+nC3x3+…+nCnxnDifferentiating, we get   n(1+x)n−1=nC1+2 nC2x+3 nC3x2+…+n nCnxn−1Putting x=-1, we get          0=nC1−2 nC2+3 nC3−…−(−1)n−1(n) nCn⇒ 0=FThus,     E=0+0=0.

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