Question

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

• A. $ {{2}^{n-1}} $
• B. $ 2\,n $
• C. $ n $
• D. $ {{2}^{n}} $

Answer 1

The Correct Option is D

Given, $ f(x)={{(1+x)}^{n}} $
On differentiating w. r. t. x, we get
$ f'(x)=n{{(1+x)}^{n-1}} $
Again, differentiating, we get
$ f''(x)=n(n-1)\,{{(1+x)}^{n-2}} $
$ \therefore $ $ {{f}^{n}}(x)=n(n-1).....\,\,\,3.2.1=n! $
$ \therefore $ $ f(0)+f'(0)+\frac{f''(0)}{2!}+....\frac{{{f}^{n}}(0)}{n!} $
$ =1+n+\frac{n(n-1)}{2!}+....+\frac{n!}{n!} $
$ {{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}{{+}^{n}}{{C}_{2}}+...{{+}^{n}}{{C}_{n}} $
$ ={{2}^{n}} $