frac{1 + frac{2^2}{2!} + frac{2^4}{3!} + frac | vedclass

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

Question

(D) Let the numerator be $N = 1 + \frac{2^2}{2!} + \frac{2^4}{3!} + \frac{2^6}{4!} + \dots \infty$. Multiplying and dividing by $2^2$,we get $N = \fra

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Let the numerator be $N = 1 + \frac{2^2}{2!} + \frac{2^4}{3!} + \frac{2^6}{4!} + \dots \infty$. Multiplying and dividing by $2^2$,we get $N = \frac{1}{2^2} \left( \frac{2^2}{1!} + \frac{2^4}{2!} + \frac{2^6}{3!} + \dots \right) = \frac{1}{4} (e^{2^2} - 1) = \frac{e^4 - 1}{4}$. Let the denominator be $D = 1 + \frac{1}{2!} + \frac{2}{3!} + \frac{2^2}{4!} + \dots \infty$. Multiplying and dividing by $2^2$,we get $D = \frac{1}{2^2} \left( 2^2 + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} + \dots \right) = \frac{1}{4} (e^2 - 1 - 2 + 2) = \frac{e^2 - 1}{4}$. Thus,the ratio is $\frac{(e^4 - 1)/4}{(e^2 - 1)/4} = \frac{e^4 - 1}{e^2 - 1} = \frac{(e^2 - 1)(e^2 + 1)}{e^2 - 1} = e^2 + 1$.

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

Similar Questions

The coefficient of $x^{6}$ in the expansion of $e^{2x}$ is:

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

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

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

The sum of the series $\frac{1^2}{1 \cdot 2!} + \frac{1^2 + 2^2}{2 \cdot 3!} + \frac{1^2 + 2^2 + 3^2}{3 \cdot 4!} + \dots + \frac{1^2 + 2^2 + \dots + n^2}{n(n + 1)!} + \dots \infty$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module