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Mathematics Question on Differentiability

If f(x)=(1+x)n,f(x)={{(1+x)}^{n}}, then the value of f(0)+f′(0)+f′ ′(0)2!+.....+f(n)(0)n!f(0)+f'(0)+\frac{f'\,'(0)}{2!}+.....+\frac{{{f}^{(n)}}(0)}{n!} is equal to

A

2n−1{{2}^{n-1}}

B

2 n2\,n

C

nn

D

2n{{2}^{n}}

Answer

2n{{2}^{n}}

Explanation

Solution

Given, f(x)=(1+x)nf(x)={{(1+x)}^{n}}
On differentiating w. r. t. x, we get
f′(x)=n(1+x)n−1f'(x)=n{{(1+x)}^{n-1}}
Again, differentiating, we get
f′′(x)=n(n−1) (1+x)n−2f''(x)=n(n-1)\,{{(1+x)}^{n-2}}
∴\therefore fn(x)=n(n−1).....   3.2.1=n!{{f}^{n}}(x)=n(n-1).....\,\,\,3.2.1=n!
∴\therefore f(0)+f′(0)+f′′(0)2!+....fn(0)n!f(0)+f'(0)+\frac{f''(0)}{2!}+....\frac{{{f}^{n}}(0)}{n!}
=1+n+n(n−1)2!+....+n!n!=1+n+\frac{n(n-1)}{2!}+....+\frac{n!}{n!}
=nC0+nC1+nC2+...+nCn{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}{{+}^{n}}{{C}_{2}}+...{{+}^{n}}{{C}_{n}}
=2n={{2}^{n}}