frac{1}{1!} + frac{1 + 2}{2!} + frac{1 + 2 + | vedclass

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

Question

(C) The $n^{th}$ term of the series is given by $T_n = \frac{1 + 2 + 2^2 + .... + 2^{n-1}}{n!}$.Using the sum of a geometric progression,$T_n = \frac{

Vedclass · Accepted Answer

$e^2 - e$

Vedclass · Answer

(C) The $n^{th}$ term of the series is given by $T_n = \frac{1 + 2 + 2^2 + .... + 2^{n-1}}{n!}$.Using the sum of a geometric progression,$T_n = \frac{2^n - 1}{(2 - 1)n!} = \frac{2^n}{n!} - \frac{1}{n!}$.Summing the series from $n=1$ to $\infty$:$S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \left( \frac{2^n}{n!} - \frac{1}{n!} \right) = \sum_{n=1}^{\infty} \frac{2^n}{n!} - \sum_{n=1}^{\infty} \frac{1}{n!}$.We know that $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + \sum_{n=1}^{\infty} \frac{x^n}{n!}$,so $\sum_{n=1}^{\infty} \frac{x^n}{n!} = e^x - 1$.For $x=2$,$\sum_{n=1}^{\infty} \frac{2^n}{n!} = e^2 - 1$.For $x=1$,$\sum_{n=1}^{\infty} \frac{1^n}{n!} = e^1 - 1 = e - 1$.Therefore,$S = (e^2 - 1) - (e - 1) = e^2 - e$.

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

Similar Questions

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

The coefficient of ${x^r}$ in the expansion of ${e^{e^x}}$ is

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

The value of $\frac{2}{3!} + \frac{4}{5!} + \frac{6}{7!} + \dots$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module