The value of 1 - log 2 + frac{(log 2)^2}{2!} | vedclass

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

Question

(B) The given series is $1 - \log 2 + \frac{(\log 2)^2}{2!} - \frac{(\log 2)^3}{3!} + \dots$ This is the expansion of the exponential series $e^x = 1

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(B) The given series is $1 - \log 2 + \frac{(\log 2)^2}{2!} - \frac{(\log 2)^3}{3!} + \dots$ This is the expansion of the exponential series $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$ where $x = -\log 2$. Substituting $x = -\log 2$ into the series,we get $e^{-\log 2}$. Using the property of logarithms,$e^{-\log 2} = e^{\log(2^{-1})} = e^{\log(1/2)}$. Since $e^{\log_e y} = y$,we have $e^{\log(1/2)} = \frac{1}{2} = 0.5$.

The value of $1 - \log 2 + \frac{(\log 2)^2}{2!} - \frac{(\log 2)^3}{3!} + \dots$ is

Similar Questions

$1 + \frac{2^3}{2!} + \frac{3^3}{3!} + \frac{4^3}{4!} + \dots \infty =$ (in $e$)

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

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

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module