1.5 + frac{2.6}{1!} + frac{3.7}{2!} + frac{4. | vedclass

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(A) The general term of the series is $T_n = \frac{(n+1)(n+4)}{n!}$ for $n \ge 1$.$T_n = \frac{n^2 + 5n + 4}{n!} = \frac{n(n-1) + 6n + 4}{n!} = \frac{

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$13e$

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(A) The general term of the series is $T_n = \frac{(n+1)(n+4)}{n!}$ for $n \ge 1$.$T_n = \frac{n^2 + 5n + 4}{n!} = \frac{n(n-1) + 6n + 4}{n!} = \frac{n(n-1)}{n!} + \frac{6n}{n!} + \frac{4}{n!}$$T_n = \frac{1}{(n-2)!} + \frac{6}{(n-1)!} + \frac{4}{n!}$Summing from $n=1$ to $\infty$:$S = \sum_{n=1}^{\infty} \frac{1}{(n-2)!} + \sum_{n=1}^{\infty} \frac{6}{(n-1)!} + \sum_{n=1}^{\infty} \frac{4}{n!}$Note that $\frac{1}{(n-2)!} = 0$ for $n=1$,so the sum starts from $n=2$:$S = (e) + 6(e) + 4(e-1) = 11e - 4 + 4 = 11e$ (Wait,re-evaluating the series terms: $1.5 = \frac{1 	imes 4}{0!}$ is not correct. Let's re-examine the series: $T_1 = 1.5 = \frac{3}{2}$,$T_2 = \frac{12}{1} = 12$,$T_3 = \frac{21}{2} = 10.5$. The general term is $T_n = \frac{(n+1)(n+4)}{n!}$).Correcting the sum: $S = \sum_{n=0}^{\infty} \frac{(n+1)(n+4)}{n!} = \sum_{n=0}^{\infty} \frac{n^2+5n+4}{n!} = \sum \frac{n(n-1)+6n+4}{n!} = e + 6e + 4e = 11e$.

$1.5 + \frac{2.6}{1!} + \frac{3.7}{2!} + \frac{4.8}{3!} + \dots$ is equal to

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