The sum of the series (1^2 + 1) cdot 1! + (2^ | vedclass

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

Question

(B) Let the $n^{th}$ term be $T_n = (n^2 + 1) \cdot n!$. We can rewrite $n^2 + 1$ as $n(n+1) - (n-1)$. So,$T_n = [n(n+1) - (n-1)] \cdot n! = n(n+1) \c

Vedclass · Accepted Answer

$n \cdot (n + 1)!$

Vedclass · Answer

(B) Let the $n^{th}$ term be $T_n = (n^2 + 1) \cdot n!$. We can rewrite $n^2 + 1$ as $n(n+1) - (n-1)$. So,$T_n = [n(n+1) - (n-1)] \cdot n! = n(n+1) \cdot n! - (n-1) \cdot n!$. $T_n = n \cdot (n+1)! - (n-1) \cdot n!$. Let $f(n) = n \cdot (n+1)!$. Then $T_n = f(n) - f(n-1)$. The sum $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} [f(k) - f(k-1)]$. This is a telescoping series: $S_n = f(n) - f(0)$. $f(n) = n \cdot (n+1)!$ and $f(0) = 0 \cdot 1! = 0$. Therefore,$S_n = n \cdot (n+1)!$.

The sum of the series $(1^2 + 1) \cdot 1! + (2^2 + 1) \cdot 2! + (3^2 + 1) \cdot 3! + \dots + (n^2 + 1) \cdot n!$ is:

Similar Questions

If $\frac{1}{2 \times 3 \times 4} + \frac{1}{3 \times 4 \times 5} + \frac{1}{4 \times 5 \times 6} + \dots + \frac{1}{100 \times 101 \times 102} = \frac{k}{101}$,then $34k$ is equal to $.....$

The sum of the infinite series $\frac{1}{3 \times 7} + \frac{1}{7 \times 11} + \frac{1}{11 \times 15} + \dots$ is

If $t_n = \frac{1}{4}(n+2)(n+3)$ for $n = 1, 2, 3, \ldots$, then $\frac{1}{t_1} + \frac{1}{t_2} + \ldots + \frac{1}{t_{2003}}$ is equal to

The sum of the infinite series $\cot ^{-1}\left(\frac{7}{4}\right)+\cot ^{-1}\left(\frac{19}{4}\right)+\cot ^{-1}\left(\frac{39}{4}\right)+\cot ^{-1}\left(\frac{67}{4}\right)+\ldots \ldots$ is :-

If $\frac{1}{2 \times 4} + \frac{1}{4 \times 6} + \frac{1}{6 \times 8} + \dots (n \text{ terms}) = \frac{k n}{4(n + 1)}$,then $k$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module