A) \[2e\]
B) \[3e\]
C) \[2e-1\]
D) None of these
Correct Answer: A
Solution :
Let \[S=1+\frac{{{2}^{2}}}{2!}+\frac{{{3}^{2}}}{3!}+\frac{{{4}^{2}}}{4!}+...\] \[\therefore \] \[{{T}_{n}}=\frac{{{n}^{2}}}{n!}=\frac{n}{(n-1)!}\] \[\Rightarrow \] \[{{T}_{n}}=\frac{n-1+1}{(n-1)!}=\frac{1}{(n-2)!}+\frac{1}{(n-1)!}\] \[\therefore \] \[{{S}_{n}}=\Sigma {{T}_{n}}=\Sigma \frac{1}{(n-2)!}+\Sigma \frac{1}{(n-1)!}\] \[=e+e\] \[=2e\]
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