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(C) The $n^{th}$ term of the series is given by $T_n = \frac{1 + 3 + 5 + \dots + (2n - 1)}{n!}$.Since the sum of the first $n$ odd numbers is $n^2$,we

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$2e$

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(C) The $n^{th}$ term of the series is given by $T_n = \frac{1 + 3 + 5 + \dots + (2n - 1)}{n!}$.Since the sum of the first $n$ odd numbers is $n^2$,we have $T_n = \frac{n^2}{n!} = \frac{n}{(n-1)!}$.We can rewrite the numerator as $n = (n-1) + 1$,so $T_n = \frac{n-1}{(n-1)!} + \frac{1}{(n-1)!} = \frac{1}{(n-2)!} + \frac{1}{(n-1)!}$ for $n \geq 2$.For $n=1$,$T_1 = 1$.Summing the series $S = \sum_{n=1}^{\infty} T_n = 1 + \sum_{n=2}^{\infty} \frac{1}{(n-2)!} + \sum_{n=1}^{\infty} \frac{1}{(n-1)!}$.$S = 1 + (1 + \frac{1}{1!} + \frac{1}{2!} + \dots) + (1 + \frac{1}{1!} + \frac{1}{2!} + \dots) = 1 + e + e = 1 + 2e$ is incorrect based on the standard series expansion.Re-evaluating: $T_n = \frac{n^2}{n!} = \frac{n}{(n-1)!} = \frac{(n-1)+1}{(n-1)!} = \frac{1}{(n-2)!} + \frac{1}{(n-1)!}$.Summing from $n=1$ to $\infty$: $T_1 = 1/1 = 1$. $T_2 = 4/2 = 2$. $T_3 = 9/6 = 1.5$.Sum $= \sum_{n=1}^{\infty} \frac{n}{(n-1)!} = \sum_{k=0}^{\infty} \frac{k+1}{k!} = \sum_{k=0}^{\infty} \frac{k}{k!} + \sum_{k=0}^{\infty} \frac{1}{k!} = \sum_{k=1}^{\infty} \frac{1}{(k-1)!} + e = e + e = 2e$.

$1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + \frac{1 + 3 + 5 + 7}{4!} + \dots \infty = $

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