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

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

Question

(D) The inner sum is a geometric series: $\sum_{n=1}^k 2^{n-1} = \frac{1(2^k - 1)}{2 - 1} = 2^k - 1$. Substituting this into the main expression: $\su

Vedclass · Accepted Answer

$e^2-e$

Vedclass · Answer

(D) The inner sum is a geometric series: $\sum_{n=1}^k 2^{n-1} = \frac{1(2^k - 1)}{2 - 1} = 2^k - 1$. Substituting this into the main expression: $\sum_{k=1}^{\infty} \frac{2^k - 1}{k!} = \sum_{k=1}^{\infty} \frac{2^k}{k!} - \sum_{k=1}^{\infty} \frac{1}{k!}$. We know the Taylor series expansion $e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} = 1 + \sum_{k=1}^{\infty} \frac{x^k}{k!}$,so $\sum_{k=1}^{\infty} \frac{x^k}{k!} = e^x - 1$. For the first part,$x = 2$: $\sum_{k=1}^{\infty} \frac{2^k}{k!} = e^2 - 1$. For the second part,$x = 1$: $\sum_{k=1}^{\infty} \frac{1^k}{k!} = e^1 - 1 = e - 1$. Subtracting these: $(e^2 - 1) - (e - 1) = e^2 - 1 - e + 1 = e^2 - e$.

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

Similar Questions

The solution of the differential equation $\frac{dy}{dx} = \frac{x - y + 3}{2(x - y) + 5}$ is

An electron initially present in an excited state of $H$ atom is further excited to another energy level by an incident photon. It releases $10$ photons while coming back to the ground state,out of which $7$ have higher energy than the incident photon. The electron was initially present in

The number of positive odd divisors of $216$ is

The time period of a particle in simple harmonic motion is $8 \ s$. At $t=0$,it is at the mean position. The ratio of the distances travelled by it in the first and second seconds is

The value of $k$ such that the lines $2x - 3y + k = 0$,$3x - 4y - 13 = 0$,and $8x - 11y - 33 = 0$ are concurrent,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module