A) \[5\,e\]
B) \[e\]
C) \[15\,e\]
D) \[2\,e\]
Correct Answer: C
Solution :
\[1+\frac{{{2}^{4}}}{2!}+\frac{{{3}^{4}}}{3!}+\frac{{{4}^{4}}}{4!}+....\infty \] \[{{T}_{n}}=\frac{{{n}^{4}}}{n!}=\frac{{{n}^{3}}}{(n-1)!}=\frac{{{n}^{3}}-1}{(n-1)!}+\frac{1}{(n-1)!}\] \[=\frac{(n-1)({{n}^{2}}+n+1)}{(n-1)!}+\frac{1}{(n-1)!}=\frac{{{n}^{2}}+n+1}{(n-2)!}+\frac{1}{(n-1)!}\] \[=\frac{{{n}^{2}}-4}{(n-2)!}+\frac{(n-2)}{(n-2)!}+\frac{7}{(n-2)!}+\frac{1}{(n-1)!}\] \[=\frac{n+2}{(n-3)!}+\frac{1}{(n-3)!}+\frac{7}{(n-2)!}+\frac{1}{(n-1)!}\] =\[\frac{1}{(n-4)!}+\frac{6}{(n-3)!}+\frac{7}{(n-2)!}+\frac{1}{(n-1)!}\] =\[\,e+6e+7e+e=15e\].
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