If n is even,then in the expansion of {left( | vedclass

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

Question

(D) We know that the series expansion of $\cosh(x)$ is $1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots = \frac{e^x + e^{-x}}{2}$.Thus,the given expressio

Vedclass · Accepted Answer

$\frac{{{2^{n - 1}}}}{{n!}}$

Vedclass · Answer

(D) We know that the series expansion of $\cosh(x)$ is $1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots = \frac{e^x + e^{-x}}{2}$.Thus,the given expression is ${\left( \frac{e^x + e^{-x}}{2} \right)^2}$.Expanding this,we get $\frac{1}{4}(e^{2x} + e^{-2x} + 2)$.Using the expansion $e^{2x} + e^{-2x} = 2 \sum_{k=0}^{\infty} \frac{(2x)^{2k}}{(2k)!} = 2 \left( 1 + \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} + \dots + \frac{(2x)^n}{n!} + \dots \right)$.Substituting this back,the expression becomes $\frac{1}{4} \left( 2 + 2 \sum_{k=0}^{\infty} \frac{(2x)^{2k}}{(2k)!} \right) = \frac{1}{2} + \frac{1}{2} \sum_{k=0}^{\infty} \frac{2^{2k} x^{2k}}{(2k)!}$.For an even $n > 0$,the coefficient of $x^n$ is $\frac{1}{2} \cdot \frac{2^n}{n!} = \frac{2^{n-1}}{n!}$.

If $n$ is even,then in the expansion of ${\left( {1 + \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}} + \dots} \right)^2}$,the coefficient of ${x^n}$ is

Similar Questions

The coefficient of ${x^2}$ in the expansion of ${\left( x - \frac{1}{2x} \right)^8}$ is

The value of $x$ in the expression $(x + x^{\log_{10} x})^5$,if the third term in the expansion is $1,000,000$.

The coefficient of $x^{24}$ in the expansion of $(1+x^2)^{12}(1+x^{12})(1+x^{24})$ is

The ratio of the coefficient of the middle term in the expansion of $(1+x)^{20}$ and the sum of the coefficients of two middle terms in the expansion of $(1+x)^{19}$ is $....$

The greatest term in the expansion of $\sqrt{3} \left( 1 + \frac{1}{\sqrt{3}} \right)^{20}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module