A) 0
B) 1
C) \[\frac{7e}{2}\]
D) \[2\,e\]
Correct Answer: C
Solution :
\[\frac{1}{0!}+\frac{1+2}{1!}+\frac{1+2+3}{2!}+....\infty \] \[{{n}^{th}}\]term \[{{T}_{n}}=\frac{1+2+3+4+.....+n}{(n-1)!}=\frac{n(n+1)}{2(n-1)!}\] \[{{T}_{n}}=\frac{1}{2}\left[ \frac{1}{(n-3)!}+\frac{4}{(n-2)!}+\frac{2}{(n-1)!} \right]\] Therefore sum\[{{S}_{\infty }}=\frac{7e}{2}\].
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