Q61 / 200 · Manipal (MET) 2012 · Mathematics

The sum of the series $1+\frac{2^2}{2!}+\frac{3^2}{3!}+\frac{4^2}{4!}+\ldots$ is

  • Correct answer
    A
    $2 e$
  • B
    $3 e$
  • C
    $2 e-1$
  • D
    None of these

Solution

$\begin{aligned} & \text { Let } S=1+\frac{2^2}{2!}+\frac{3^2}{3!}+\frac{4^2}{4!}+\ldots \\ & \therefore \quad T_n=\frac{n^2}{n!}=\frac{n}{(n-1)!} \\ & \Rightarrow T_n=\frac{n-1+1}{(n-1)!}=\frac{1}{(n-2)!}+\frac{1}{(n-1)!} \\ & \therefore \quad S_n=\Sigma T_n=\Sigma \frac{1}{(n-2)!}+\Sigma \frac{1}{(n-1)!} \\ & =e+e \\ & =2 e \end{aligned}$

Hence, the correct answer is option: 'A'.

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