1 + frac{2^2}{1!} + frac{3^2}{2!} + frac{4^2} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The general term of the series is $T_n = \frac{n^2}{(n-1)!}$ for $n \ge 1$.We can rewrite the numerator as $n^2 = n(n-1) + n = n(n-1) + (n-1) + 1$

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) The general term of the series is $T_n = \frac{n^2}{(n-1)!}$ for $n \ge 1$.We can rewrite the numerator as $n^2 = n(n-1) + n = n(n-1) + (n-1) + 1$.So,$T_n = \frac{n(n-1)}{(n-1)!} + \frac{n-1}{(n-1)!} + \frac{1}{(n-1)!} = \frac{1}{(n-3)!} + \frac{1}{(n-2)!} + \frac{1}{(n-1)!}$ (for $n \ge 3$).Summing the series $S = \sum_{n=1}^{\infty} T_n$:$S = \frac{1^2}{0!} + \frac{2^2}{1!} + \sum_{n=3}^{\infty} \left( \frac{1}{(n-3)!} + \frac{1}{(n-2)!} + \frac{1}{(n-1)!} \right)$$S = 1 + 4 + (e + e + (e-1)) = 5 + 3e - 1 = 4 + 3e$ (Wait,re-evaluating).Correct expansion: $T_n = \frac{(n-1+1)^2}{(n-1)!} = \frac{(n-1)^2 + 2(n-1) + 1}{(n-1)!} = \frac{(n-1)(n-2) + (n-1) + 2(n-1) + 1}{(n-1)!} = \frac{1}{(n-3)!} + \frac{3}{(n-2)!} + \frac{1}{(n-1)!}$.Summing from $n=1$ to $\infty$: $S = e + 3e + e = 5e$.

$1 + \frac{2^2}{1!} + \frac{3^2}{2!} + \frac{4^2}{3!} + \dots \infty = $ (in $e$)

Similar Questions

$\sum_{n=1}^{\infty} \frac{2n^2+n+1}{n!}$ is equal to

The sum of the series $\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots$ is

The expression $\begin{aligned} & 1+x \log _e a+\frac{x^2}{2 !}\left(\log _e a\right)^2+\frac{x^3}{3 !}\left(\log _e a\right)^3+\ldots \end{aligned}$ for $a>0, x \in R$ is equal to:

$1 + \frac{1}{3!} + \frac{1}{5!} + \frac{1}{7!} + \dots \infty = $

$1 + \frac{4^2}{3!} + \frac{4^4}{5!} + \dots \infty = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module