Question

If ${e^e}^x=a_0+a_1{x}^2+a_2{x}^2+...$, then

• A. $a_0=1$
• B. $a_0=e$
• C. $a_0=e^e$
• D. $a_0=e^2$

Answer 1

The correct option is B

$a_0=e$

Finding the value:

$\begin{array}{rcl}{e}^{e^x}& =& 1+\frac{e^x}{1!}+\frac{e^{2x}}{2!}+\frac{e^{3x}}{3!}+...\\ & =& 1+\frac{1}{1!}(1+x+\frac{x^2}{2!}+\dots .)+\frac{1}{2!}(1+\frac{2x}{1!}+\frac{{\left(2x\right)}^2}{2!}+\dots .)+...\\ & =& \left(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots \right)+x\left(\frac{1}{1!}+\frac{2}{2!}+\dots \right)+\dots .\end{array}$

So,

$a_0=e$

Hence, option (B) is the answer.