Let S denote the sum of the infinite series 1 | vedclass

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(C) The general term $T_n$ of the series $1, 8, 21, 40, 65, \ldots$ in the numerator can be found using the method of differences. The first differenc

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$8 < S < 12$

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(C) The general term $T_n$ of the series $1, 8, 21, 40, 65, \ldots$ in the numerator can be found using the method of differences. The first differences are $7, 13, 19, 25, \ldots$, which form an arithmetic progression with common difference $6$. Thus, $T_n = an^2 + bn + c$. For $n=1, T_1 = a+b+c = 1$. For $n=2, T_2 = 4a+2b+c = 8$. For $n=3, T_3 = 9a+3b+c = 21$. Solving these, we get $a=3, b=-2, c=0$. So, $T_n = 3n^2 - 2n = n(3n-2)$. The series is $S = \sum_{n=1}^{\infty} \frac{n(3n-2)}{n!} = \sum_{n=1}^{\infty} \frac{3n^2-2n}{n!}$. Since the first term is $1$ (for $n=1$), we can write $S = \sum_{n=1}^{\infty} \frac{3n-2}{(n-1)!}$. Let $k = n-1$, then $n = k+1$. $S = \sum_{k=0}^{\infty} \frac{3(k+1)-2}{k!} = \sum_{k=0}^{\infty} \frac{3k+1}{k!} = 3\sum_{k=1}^{\infty} \frac{k}{k!} + \sum_{k=0}^{\infty} \frac{1}{k!} = 3\sum_{k=1}^{\infty} \frac{1}{(k-1)!} + e = 3e + e = 4e$. Since $2 Therefore, $8 < S < 12$.

Let $S$ denote the sum of the infinite series $1+\frac{8}{2!}+\frac{21}{3!}+\frac{40}{4!}+\frac{65}{5!}+\ldots$. Then

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