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

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(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

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$n \cdot (n + 1)!$

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(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:

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