sum_{k=1}^{infty} frac{1}{k !} left(sum_{n=1} | vedclass

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

Question

(D) The given sum is $S = \sum_{k=1}^{\infty} \frac{1}{k !} \left(\sum_{n=1}^k 2^{n-1}\right)$.Inside the bracket,we have a geometric series with $a=1

Vedclass · Accepted Answer

$e^2-e$

Vedclass · Answer

(D) The given sum is $S = \sum_{k=1}^{\infty} \frac{1}{k !} \left(\sum_{n=1}^k 2^{n-1}\right)$.Inside the bracket,we have a geometric series with $a=1$,$r=2$,and $k$ terms,so $\sum_{n=1}^k 2^{n-1} = \frac{1(2^k-1)}{2-1} = 2^k-1$.Substituting this into the expression,we get $S = \sum_{k=1}^{\infty} \frac{2^k-1}{k !}$.This can be split into two sums: $S = \sum_{k=1}^{\infty} \frac{2^k}{k !} - \sum_{k=1}^{\infty} \frac{1}{k !}$.We know that $e^x = \sum_{k=0}^{\infty} \frac{x^k}{k !} = 1 + \sum_{k=1}^{\infty} \frac{x^k}{k !}$,so $\sum_{k=1}^{\infty} \frac{x^k}{k !} = e^x - 1$.For the first sum,$x=2$,so $\sum_{k=1}^{\infty} \frac{2^k}{k !} = e^2 - 1$.For the second sum,$x=1$,so $\sum_{k=1}^{\infty} \frac{1^k}{k !} = e^1 - 1 = e - 1$.Thus,$S = (e^2 - 1) - (e - 1) = e^2 - e$.

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

Similar Questions

${\left[ {1 + \frac{1}{{2!}} + \frac{1}{{4!}} + \dots \infty } \right]^2} - {\left[ {1 + \frac{1}{{3!}} + \frac{1}{{5!}} + \dots \infty } \right]^2} = $

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

In the expansion of $(1 + x + x^2)e^{-x}$,the coefficient of $x^2$ is

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

The coefficient of $x^n$ in $\frac{1-2x}{e^x}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module