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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(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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