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Question

If f(x)=xn+4, then f(1)+f(1)1!+f′′(1)2!+f′′′(1)3!++fn(1)n!=

A
2n1
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B
2n+4
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C
2n
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D
1+11!+12!++1n!
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Solution

The correct option is B 2n+4
f(x)=1+4
f(x)=n(x)n1
f(1)=n
f′′(x)=n(x)n1xn2
f′′(1)=n(n1)
f′′′(x)=n(n1)(n2)xn2
f′′′(1)=n(n1)(n2)
=f(1)+f(1)1!+f(1)2!+cf′′′(1)3!...........fn(1)n!
=4+1+n+n(n1)2!+n(n2)(n1)3!+n!n!
=4+nC0+nC1+nC2+nC3nCn
=4+2n

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