$Đặt$ $tử$ $số$ $là$ $A$,$mẫu$ $số$ $là$ $C$
⇒ $A$ $=$ $1$ + $2$ + $2^2$ + $2^3$ +....+ $2^{2019}$
$2A$ $=$ $2$ + $2^2$ + $2^3$ +...+ $2^{2020}$
$2A-A$ $=$$2$+$2^2$+$2^3$+...+$2^{2020}$-($1$+$2$+$2^2$+$2^3$+....+$2^{2019}$)
$A$ $=$ $2$+$2^2$+$2^3$+...+$2^{2020}$-$1$-$2$-$2^2$-$2^3$-...-$2^{2019}$
$A$ $=$ $2^{2020}$-$1$
$Vậy$ $ta$ $có$ $phân$ $số$ $là :$
$\frac{A}{C}$ =$\frac{2^{2020}-1}{1-2^{2020}}$ = - $\frac{1-2^{2020}}{1-2^{2020}}$ = $-1$
