Question

$e^2-e=1+\frac{1+2}{a!}+\frac{1+2+2^2}{3!}+\frac{1+2+2^2+2^3}{c!}+...$ Find $c÷a$

Answer 1

We know, $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+................$
$\therefore e^2=1+2+\frac{2^2}{2!}+\frac{2^3}{3!}+................$
and $e=1+1+\frac{1}{2!}+\frac{1}{3!}+................$
$\therefore e^2-e=0+1+\frac{2^2-1}{2!}+\frac{2^3-1}{3!}+................$
$=1+\frac{1+2}{2!}+\frac{1+2+2^2}{3!}+\frac{1+2+2^2+2^3}{3!}+................$ $,\because [2^n-1=1+2+2^2+........{.2}^{n-1}]$
$\Rightarrow c=4,a=2\therefore c÷a=2$