Q59 / 170 · Manipal (MET) 2021 · Mathematics

$1+\frac{3}{1!}+\frac{5}{2!}+\frac{7}{3!}+\ldots \infty$ is

  • A
    5
  • B
    $2 e$
  • Correct answer
    C
    $3 e$
  • D
    None of these

Solution

Let $T_n$ be the $n$th term of a given series: $\therefore \quad T_n=\frac{(2 n-1)}{(n-1)!} ; n=1,2,3, \ldots, \infty$ $\begin{aligned} \Rightarrow \quad T_n & =\frac{2 n}{(n-1)!}-\frac{1}{(n-1)!} \\ & =2\left[\frac{(n-1)}{(n-1)!}+\frac{1}{(n-1)!}\right]-\frac{1}{(n-1)!} \\ & =\frac{2}{(n-2)!}+\frac{1}{(n-1)!} \\ \therefore \sum_{n=1}^{\infty} T_n & =\sum_{n=1}^{\infty} \frac{2}{(n-2)!}+\sum_{n=1}^{\infty} \frac{1}{(n-1)!} \\ & =2 e+e=3 e\end{aligned}$

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

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