{left[ {1 + frac{1}{{2!}} + frac{1}{{4!}} + d | vedclass

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

Question

(B) We know that the expansion of $e^x$ is $1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \infty$. For $x = 1$,$e = 1 + 1 + \frac{1}{2!} + \frac{1}{

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) We know that the expansion of $e^x$ is $1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \infty$. For $x = 1$,$e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots \infty$. For $x = -1$,$e^{-1} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots \infty$. Adding these,$e + e^{-1} = 2(1 + \frac{1}{2!} + \frac{1}{4!} + \dots \infty)$,so $1 + \frac{1}{2!} + \frac{1}{4!} + \dots \infty = \frac{e + e^{-1}}{2}$. Subtracting these,$e - e^{-1} = 2(1 + \frac{1}{3!} + \frac{1}{5!} + \dots \infty)$,so $1 + \frac{1}{3!} + \frac{1}{5!} + \dots \infty = \frac{e - e^{-1}}{2}$. Substituting these into the expression: ${\left( \frac{e + e^{-1}}{2} ight)^2} - {\left( \frac{e - e^{-1}}{2} ight)^2} = \frac{1}{4} \left[ (e + e^{-1})^2 - (e - e^{-1})^2 ight]$. Using the identity $(a+b)^2 - (a-b)^2 = 4ab$,we get $\frac{1}{4} 	imes 4 	imes e 	imes e^{-1} = 1$.

${\left[ {1 + \frac{1}{{2!}} + \frac{1}{{4!}} + \dots \infty } \right]^2} - {\left[ {1 + \frac{1}{{3!}} + \frac{1}{{5!}} + \dots \infty } \right]^2} = $

Similar Questions

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

$\sum_{k=1}^{\infty} \frac{1}{k !} \left(\sum_{n=1}^k 2^{n-1}\right)$ is equal to

Which one of the following is not correct for the features of the exponential function given by $f(x) = b^{x}$ where $b > 1$?

The sum of the series $1 + \frac{1}{4 \cdot 2!} + \frac{1}{16 \cdot 4!} + \frac{1}{64 \cdot 6!} + \dots$ to infinity is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module