tính B= 1+2+2^2+2^3+….+2^2019/1-2^2020

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1. $Đặt$ $tử$ $số$ $là$ $A$,$mẫu$ $số$ $là$ $C$

   ⇒ $A$ $=$ $1$ + $2$ + $2^2$ + $2^3$ +….+ $2^{2019}$

   $2A$ $=$ $2$ + $2^2$ + $2^3$ +…+ $2^{2020}$

   $2A-A$ $=$$2$+$2^2$+$2^3$+…+$2^{2020}$-($1$+$2$+$2^2$+$2^3$+….+$2^{2019}$)

   $A$ $=$ $2$+$2^2$+$2^3$+…+$2^{2020}$-$1$-$2$-$2^2$-$2^3$-…-$2^{2019}$

   $A$ $=$ $2^{2020}$-$1$

   $Vậy$ $ta$ $có$ $phân$ $số$ $là :$

   $\frac{A}{C}$ =$\frac{2^{2020}-1}{1-2^{2020}}$ = – $\frac{1-2^{2020}}{1-2^{2020}}$ = $-1$

    

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2. Đặt tử số là 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 :

   B=$\frac{2^{2020}-1}{1-2^{2020}}$ = – $\frac{1-2^{2020}}{1-2^{2020}}$ = $-1$

    

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