frac{2}{1!}(log_e 2) + frac{2^2}{2!}(log_e 2) | vedclass

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

Question

(B) The given series is of the form $\sum_{n=1}^{\infty} \frac{x^n}{n!}$,where $x = 2 \log_e 2$.We know that the expansion of $e^x$ is $1 + \frac{x}{1

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) The given series is of the form $\sum_{n=1}^{\infty} \frac{x^n}{n!}$,where $x = 2 \log_e 2$.We know that the expansion of $e^x$ is $1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \infty$.Therefore,$\sum_{n=1}^{\infty} \frac{x^n}{n!} = e^x - 1$.Substituting $x = 2 \log_e 2 = \log_e(2^2) = \log_e 4$,we get:Sum $= e^{\log_e 4} - 1$.Since $e^{\log_e a} = a$,the sum is $4 - 1 = 3$.

$\frac{2}{1!}(\log_e 2) + \frac{2^2}{2!}(\log_e 2)^2 + \frac{2^3}{3!}(\log_e 2)^3 + \dots \infty = $

Similar Questions

The coefficient of ${x^r}$ in the expansion of ${e^{e^x}}$ is

$\frac{2}{1!} + \frac{4}{3!} + \frac{6}{5!} + \frac{8}{7!} + \dots \infty = $

In the expansion of $\frac{e^{5x} + e^x}{e^{3x}}$,the coefficient of $x^4$ is

$\frac{2}{2!} + \frac{2+4}{3!} + \frac{2+4+6}{4!} + \dots$ is equal to

$\frac{2}{3!} + \frac{4}{5!} + \frac{6}{7!} + \dots \infty = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module