The sum of the series $1 + \frac{3}{2!} + \frac{7}{3!} + \frac{15}{4!} + \dots \text{to } \infty$ is

  • A
    $e(e + 1)$
  • B
    $e(1 - e)$
  • C
    $3e - 1$
  • D
    $e(e - 1)$

Explore More

Similar Questions

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

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

$1 + \frac{a - b}{a} + \frac{1}{2!} \left( \frac{a - b}{a} \right)^2 + \frac{1}{3!} \left( \frac{a - b}{a} \right)^3 + \dots \infty = $

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

The sum of the infinite series $1 + 2 + \frac{1}{2!} + \frac{2}{3!} + \frac{1}{4!} + \frac{2}{5!} + \dots$ is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo