A) \[\frac{n(n+1)}{2}\]
B) \[{{2}^{n}}\]
C) \[{{2}^{(n+1)}}\]
D) \[{{2}^{(n-1)}}\]
Correct Answer: B
Solution :
Given, \[f(x)={{x}^{n}},\,f(1)=1\] \[f'(x)=n{{x}^{n-1}},\,f'(1)=n\] \[f''(x)=n(n-1){{x}^{n-2}},f''(1)=n(n-1)\] \[f'''(x)=n(n-1)\,(n-2){{x}^{n-3}},\] \[f'''\,(1)=n\,(n-1)\,(n-2)\] \[\begin{matrix} . \\ . \\ . \\ \end{matrix}\] \[\begin{matrix} . \\ . \\ . \\ \end{matrix}\] \[\begin{matrix} . \\ . \\ . \\ \end{matrix}\] \[\begin{matrix} . \\ . \\ . \\ \end{matrix}\] \[\begin{matrix} . \\ . \\ . \\ \end{matrix}\] \[\begin{matrix} . \\ . \\ . \\ \end{matrix}\] \[\begin{matrix} . \\ . \\ . \\ \end{matrix}\] \[\begin{matrix} . \\ . \\ . \\ \end{matrix}\] \[\begin{matrix} . \\ . \\ . \\ \end{matrix}\] \[{{f}^{n}}(x)=n(n-1)(n-2)....1.{{x}^{n-n}},\,{{f}^{n}}(1)=n!\] ?..(i) Now, \[f(1)+\frac{f'(1)}{1!}+\frac{f''(1)}{2!}+....+\frac{{{f}^{n}}(1)}{n!}\] \[=1+\frac{n}{1!}+\frac{n(n-1)}{2!}+\frac{n(n-1)(n-2)}{3!}+.....+\frac{n!}{n!}\] \[={{(1+1)}^{n}}\] \[={{2}^{n}}\]
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