Q133 / 200 · WBJEE 2014 · Mathematics

Let $t_{n}$ denotes the nth term of the infinite series $\frac{1}{1 !}+\frac{10}{2 !}+\frac{21}{3 !}+\frac{34}{4 !}+\frac{49}{5 !}+\ldots .$ Then $\lim _{n \rightarrow \infty} t_{n}$ is

  • A
    $e$
  • Correct answer
    B
    $0$
  • C
    $e^{2}$
  • D
    1

Solution

Let $\quad S=1+10+21+34+49+\ldots+t_{n}^{\prime}$ and $S=1+10+21+34+\ldots+t_{n}^{\prime}$ $0=1+9+11+13+15+\ldots-t_{n}^{\prime}$ $$ \begin{aligned} \Rightarrow t_{n}^{\prime} &=1+[9+11+13+15+\ldots(n-1) \text { term }] \\ &=1+\left[\frac{n-1}{2}\{2 \times 9+(n-2) 2\}\right] \\ &=1+(n-1)[9+n-2] \\ &=1+(n-1)(n+7) \\ \therefore \quad t_{n} &=\frac{t^{\prime}}{n !}=\frac{1+(n-1)(n+7)}{n !} \\ &=\frac{1+(n-1)(n+7)}{n !} \\ &=\frac{1+n^{2}+6 n-7}{n !} \\ &=\frac{n}{(n-1) !}+\frac{6}{(n-1) !}-\frac{n^{2}+6 n-6}{n !} \\ &=\frac{n-1+1}{(n-1) !}+\frac{6}{(n-1) !}-\frac{1}{n !} \\ &=\frac{1}{(n-2) !}+\frac{7}{(n-1) !}-\frac{1}{n !} \\ &=\frac{1}{n \rightarrow \infty} t_{n}=\lim _{n \rightarrow \infty}\left[\frac{1}{(n-2) !}+\frac{7}{(n-1) !}-\frac{1}{n !}\right]=0 \end{aligned} $$

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

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