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 sum of the infinite series $1+\frac{1}{2!}+\frac{1 \cdot 3}{4!}+\frac{1 \cdot 3 \cdot 5}{6!}+\dots$ is

The value of $\sum\limits_{n = 1}^\infty {\frac{{^n{C_0} + ... + ^n{C_n}}}{{^n{P_n}}}} $ is

Find the sum to infinity of the series $\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$

The coefficient of $x^3$ in the expansion of $3^x$ is

The coefficient of $x^k$ in the expansion of $\frac{1-2x-x^2}{e^{-x}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module