1 + frac{3}{1!} + frac{5}{2!} + frac{7}{3!} + | vedclass

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(C) The general term of the series is $T_n = \frac{2n - 1}{(n - 1)!}$ for $n \geq 1$.We can rewrite the general term as:$T_n = \frac{2(n - 1) + 1}{(n

Vedclass · Accepted Answer

$3e$

Vedclass · Answer

(C) The general term of the series is $T_n = \frac{2n - 1}{(n - 1)!}$ for $n \geq 1$.We can rewrite the general term as:$T_n = \frac{2(n - 1) + 1}{(n - 1)!} = \frac{2(n - 1)}{(n - 1)!} + \frac{1}{(n - 1)!} = \frac{2}{(n - 2)!} + \frac{1}{(n - 1)!}$.The sum of the series is $\sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \frac{2}{(n - 2)!} + \sum_{n=1}^{\infty} \frac{1}{(n - 1)!}$.Note that for $n=1$,$\frac{2}{(n-2)!}$ is defined as $0$ (since the factorial of a negative integer is infinite in the denominator).Thus,the sum is $2 \sum_{n=2}^{\infty} \frac{1}{(n - 2)!} + \sum_{n=1}^{\infty} \frac{1}{(n - 1)!} = 2e + e = 3e$.

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

Similar Questions

$\frac{1}{2!} + \frac{1+2}{3!} + \frac{1+2+3}{4!} + \ldots$ is equal to :

$\frac{1 + \frac{2^2}{2!} + \frac{2^4}{3!} + \frac{2^6}{4!} + \dots \infty}{1 + \frac{1}{2!} + \frac{2}{3!} + \frac{2^2}{4!} + \dots \infty} = $

$\sum_{n=1}^{\infty} \frac{2n^2+n+1}{n!}$ is equal to

$3 + \frac{5}{1!} + \frac{7}{2!} + \frac{9}{3!} + \dots \infty = $

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

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