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Question

If f(x)=(1+x)n, then the value of f(0)+f(0)+f′′(0)2!+.....+fn(0)n! is

A
n
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B
2n
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C
2n1
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D
None of these
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Solution

The correct option is B 2n
f(x)=n(1+x)n1,f"(x)=n(n1)(1+x)n2
fn(x)=n!,fn(0)=n!
1+n+n(n1)2!+....+n!n!
=nC0+nC1+nC2+.....+nCn=2n

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