Question

If $\frac{e^x+e^{5x}}{e^{3x}}=a_0+a_1{x}^{}+a_2{x}^2+a_3{x}^3....$ then the value of $2a_1+2^3{a}_3+2^5{a}_5+....$ is

• A. $e^2+e^{-2}$
• B. $e^4-e^{-4}$
• C. $e^4+e^{-4}$
• D. $0$

Answer 1

Answer: (D)

$0$

Let $\frac{e^x+e^{5x}}{e^{3x}}=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+....$

$=\frac{e^x}{e^{3x}}+\frac{e^x}{e^{5x}}=a_0+a_{1}x+a_2{x}^2+...$

$=e^{-2x}+e^{2x}=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+....e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....;e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...$

$e^{-2x}+e^{2x}=2[1+\frac{{\left(2x\right)}^2}{2!}-\frac{{\left(2x\right)}^4}{4!}+...]=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+....$

by comparing the coefficient

$a_1=a_3=a_5=...=0$

Hence, $2a_1+2^{3}a+2^5{a}_5+...=0$