The coefficient of $x^r$ in the expansion of $1 + \frac{a + bx}{1!} + \frac{(a + bx)^2}{2!} + \dots + \frac{(a + bx)^n}{n!} + \dots$ is

  • A
    $\frac{(a + b)^r}{r!}$
  • B
    $\frac{b^r}{r!}$
  • C
    $\frac{e^a b^r}{r!}$
  • D
    $e^{a + b^r}$

Explore More

Similar Questions

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

$\frac{1 + \frac{2^2}{2!} + \frac{2^4}{3!} + \frac{2^6}{4!} + \dots \infty}{1 + \frac{1}{2!} + \frac{2}{3!} + \frac{2^2}{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 = $

Let $\sum_{n=0}^{\infty} \frac{n^3((2n)!) + (2n-1)(n!)}{(n!)((2n)!)} = ae + \frac{b}{e} + c$,where $a, b, c \in \mathbb{Z}$ and $e = \sum_{n=0}^{\infty} \frac{1}{n!}$. Then $a^2 - b + c$ is equal to $................$.

Find the sum of the series $\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots \infty$.

Difficult
View Solution

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