$1 + \frac{a - bx}{1!} + \frac{(a - bx)^2}{2!} + \frac{(a - bx)^3}{3!} + \dots \infty = $
- A$e^{a - bx}$
- B$e^{a - bx} - 1$
- C$1 + a \log_e(a - bx)$
- D$e^{-bx}$
Explore More
Similar Questions
The sum of the series $\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots$ is
MediumAIEEE 2004
View Solution$1+\frac{1+2}{2 !}+\frac{1+2+2^2}{3 !}+\ldots$ is equal to
$\frac{1^2 \cdot 2}{1!} + \frac{2^2 \cdot 3}{2!} + \frac{3^2 \cdot 4}{3!} + \dots \infty = $ (in $e$)
Medium
View SolutionThe value of $\sqrt{e}$ is approximately:
Medium
View Solution$\frac{x^2 - y^2}{1!} + \frac{x^4 - y^4}{2!} + \frac{x^6 - y^6}{3!} + \dots \infty = $
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo