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

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(D) The general term of the series is $T_n = \frac{n^3}{n!} = \frac{n^2}{(n-1)!}$.We can write $n^2 = n(n-1) + n = n(n-1) + (n-1) + 1$.Thus,$T_n = \fr

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) The general term of the series is $T_n = \frac{n^3}{n!} = \frac{n^2}{(n-1)!}$.We can write $n^2 = n(n-1) + n = n(n-1) + (n-1) + 1$.Thus,$T_n = \frac{n(n-1)}{(n-1)!} + \frac{n-1}{(n-1)!} + \frac{1}{(n-1)!} = \frac{1}{(n-3)!} + \frac{1}{(n-2)!} + \frac{1}{(n-1)!}$.Summing from $n=1$ to $\infty$:$S = \sum_{n=1}^{\infty} \frac{1}{(n-3)!} + \sum_{n=1}^{\infty} \frac{1}{(n-2)!} + \sum_{n=1}^{\infty} \frac{1}{(n-1)!}$.Note that $\frac{1}{k!} = 0$ for $k $S = (0 + 0 + 1 + \frac{1}{1!} + \dots) + (0 + 1 + \frac{1}{1!} + \dots) + (1 + \frac{1}{1!} + \dots) = e + e + e = 3e$ is incorrect based on the standard expansion.Re-evaluating: $T_n = \frac{n^2}{(n-1)!} = \frac{n(n-1)+n}{(n-1)!} = \frac{1}{(n-3)!} + \frac{3}{(n-2)!} + \frac{1}{(n-1)!}$.$S = e + 3e + e = 5e$.

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

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