Question

If $\frac{e-1}{e+1}=\frac{\frac{1}{a!}+\frac{1}{b!}+\frac{1}{6!}+...}{\frac{1}{1!}+\frac{1}{c!}+\frac{1}{d!}+...}$ then find $a+b-c+d$

Answer 1

We know, $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...........$
$\therefore e=1+1+\frac{1}{2!}+\frac{1}{3!}+...........\left(i\right)$
and $\therefore e^{-1}=1-1+\frac{1}{2!}-\frac{1}{3!}+...........\left(ii\right)$
Now $\frac{e-1}{e+1}=\frac{e-1}{e+1}.\frac{e-1}{e-1}=\frac{e^2+1-2e}{e^2-1}$
$=\frac{e+e^{-1}-2}{e-e^{-1}}=\frac{\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+........}{\frac{1}{1!}+\frac{1}{3!}+\frac{1}{5!}+........}$, using $\left(i\right)$ and $\left(ii\right)$