1+frac{1+2}{2 !}+frac{1+2+2^2}{3 !}+ldots is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The general term of the series is $T_n = \frac{1+2+2^2+\ldots+2^{n-1}}{n !}$ for $n \geq 1$.Using the sum of a geometric progression, $T_n = \frac

Vedclass · Accepted Answer

$e^2-e$

Vedclass · Answer

(D) The general term of the series is $T_n = \frac{1+2+2^2+\ldots+2^{n-1}}{n !}$ for $n \geq 1$.Using the sum of a geometric progression, $T_n = \frac{2^n-1}{n !}$.We can rewrite this as $T_n = \frac{2^n}{n !} - \frac{1}{n !}$.The sum of the series is $S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \frac{2^n}{n !} - \sum_{n=1}^{\infty} \frac{1}{n !}$.Recall the expansion $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n !} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \ldots$.Thus, $\sum_{n=1}^{\infty} \frac{2^n}{n !} = e^2 - 1$ and $\sum_{n=1}^{\infty} \frac{1}{n !} = e - 1$.Substituting these values, $S = (e^2 - 1) - (e - 1) = e^2 - e$.

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

Similar Questions

In the expansion of $\frac{1 + x}{1!} + \frac{(1 + x)^2}{2!} + \frac{(1 + x)^3}{3!} + \dots$,the coefficient of $x^n$ will be

If $\cosh (x-\log 3)=\sinh x$,then $x=$

The coefficient of $x^n$ in the expansion of $\frac{e^{7x}+e^x}{e^{3x}}$ is

The sum of the series $\frac{1}{2 !} + \frac{1+2}{3 !} + \frac{1+2+3}{4 !} + \ldots$ is equal to :

$\sum_{k=1}^{\infty} \frac{1}{k !} \left(\sum_{n=1}^k 2^{n-1}\right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module