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Question

If f(x)=(1+x)n, then the value of f(0)+f(0)+f′′(0)2!+...+f′′(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(0)=1, f(x)=n(1+x)n1, f′′(x)=n(n1)(1+x)n2, ..., fn(x)=n(n1)...1(1+x)nn.
So f(0)=n, f′′(0)=n(n1), ..., fn(0)=n!.
Hence the given expression is equal to
1+n+n(n1)2!+...+n!n!=nC0+nC1+nC2+...+nCn=2n

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