A) \[{{2}^{n-1}}\]
B) \[2\,n\]
C) \[n\]
D) \[{{2}^{n}}\]
Correct Answer: D
Solution :
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 \[f''(x)=n(n-1)\,{{(1+x)}^{n-2}}\] \[\therefore \] \[{{f}^{n}}(x)=n(n-1).....\,\,\,3.2.1=n!\] \[\therefore \] \[f(0)+f'(0)+\frac{f''(0)}{2!}+....\frac{{{f}^{n}}(0)}{n!}\] \[=1+n+\frac{n(n-1)}{2!}+....+\frac{n!}{n!}\] \[{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}{{+}^{n}}{{C}_{2}}+...{{+}^{n}}{{C}_{n}}\] \[={{2}^{n}}\]
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