The sum 1+frac{1+3}{2!}+frac{1+3+5}{3!}+frac{ | vedclass

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(D) The $r$-th term of the series is $T_r = \frac{1+3+5+\ldots+(2r-1)}{r!}$.Since the sum of the first $r$ odd numbers is $r^2$,we have $T_r = \frac{r

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(D) The $r$-th term of the series is $T_r = \frac{1+3+5+\ldots+(2r-1)}{r!}$.Since the sum of the first $r$ odd numbers is $r^2$,we have $T_r = \frac{r^2}{r!} = \frac{r}{(r-1)!}$.We can write $r$ as $(r-1+1)$,so $T_r = \frac{r-1+1}{(r-1)!} = \frac{r-1}{(r-1)!} + \frac{1}{(r-1)!} = \frac{1}{(r-2)!} + \frac{1}{(r-1)!}$ (for $r \ge 2$).The sum $S = \sum_{r=1}^{\infty} T_r = T_1 + \sum_{r=2}^{\infty} \left( \frac{1}{(r-2)!} + \frac{1}{(r-1)!} \right)$.$S = 1 + \sum_{r=2}^{\infty} \frac{1}{(r-2)!} + \sum_{r=2}^{\infty} \frac{1}{(r-1)!}$.$S = 1 + (\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \ldots) + (\frac{1}{1!} + \frac{1}{2!} + \ldots) = 1 + e + (e-1) = 2e$.

The sum $1+\frac{1+3}{2!}+\frac{1+3+5}{3!}+\frac{1+3+5+7}{4!}+\ldots$ up to $\infty$ terms is equal to (in $e$)

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