Question

Consider the function $g\left(x\right)$ defined as $g\left(x\right)\cdot (x^{(2^{2008}-1)}-1)=(x+1)(x^2+1)(x^4+1)....(x^{2^{2007}}+1)-1$. The value of $g\left(2\right)$ equals to ....

Answer 1

$RHS=\frac{(x-1)(x+1)(x^2+1)(x^4+1)....(x^{2^{2007}}+1)}{(x-1)}-1$
$=\frac{(x^2-1)(x^2+1)....(x^{2^{2007}}+1)}{(x-1)}-1$
$=\frac{(x^{2^2}-1)(x^{2^2}+1)....(x^{2^{2007}}+1)}{(x-1)}-1$
Hence, $g\left(x\right)\cdot (x^{(2^{2008}-1)}-1)=\frac{(x^{2^{2008}}-1)-(x-1)}{(x-1)}$
$\Rightarrow g\left(x\right)=\frac{x}{x-1}$
$\therefore g\left(2\right)=2$
Ans: 2