A) \[{{2}^{n}}\]
B) \[{{2}^{n-1}}\]
C) 0
D) 1
Correct Answer: C
Solution :
Since, \[f(x)={{x}^{n}}\] \[\Rightarrow \] \[f'(x)=n{{x}^{n-1}}\Rightarrow \,f'(1)=n\] \[f''(x)=n(n-1)\,{{x}^{n-2}}\Rightarrow f''(1)=n\,(n-1)\] ????????. ????????. \[f''(x)=n(n-1)\,(n-2)\,....\,\,2.1\] \[\Rightarrow \] \[f''(1)=n\,(n-1)\,(n-2)\,....\,2.1\] We have, \[f(1)-\frac{f'(1)}{1!}+\frac{f''(1)}{2!}-\frac{f'''(1)}{3!}+....+\frac{{{(-1)}^{n}}{{f}^{n}}(1)}{n!}\] \[=1-\frac{n}{1!}+\frac{n(n-1)}{2!}-\frac{n(n-1)\,(n-2)}{3!}+...\] \[+\frac{{{(-1)}^{n}}n(n-1)\,(n-2)\,....\,2.1}{n!}\] \[={{(1-1)}^{n}}=0\]You need to login to perform this action.
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