Câu hỏi của Phạm Thị Thùy Diễm

Question

Cho A=1+2+2^2+2^3+...+2^2008, B=200^9. Tính B-A.

Giúp tôi!!!

Nguyễn Đắc Định · Accepted Answer

Ta có : A=$2^0+2^1+2^2+2^3+...+2^{2008}$

$\Rightarrow2A=2^1+2^2+2^3+2^4+...+2^{2009}$

$\Rightarrow2A-A=$( $2^1+2^2+2^3+2^4+...+2^{2009}$ )-($2^0+2^1+2^2+2^3+...+2^{2008}$)

$\Rightarrow$   A= 2$^{2009}$-1

$\Rightarrow$B-A = $2^{2009}-\left(2^{2009}-1\right)=2^{2009}-2^{2009}+1=1$Câu trả lời: Ta có : A=$2^0+2^1+2^2+2^3+...+2^{2008}$

$\Rightarrow2A=2^1+2^2+2^3+2^4+...+2^{2009}$

$\Rightarrow2A-A=$( $2^1+2^2+2^3+2^4+...+2^{2009}$ )-($2^0+2^1+2^2+2^3+...+2^{2008}$)

$\Rightarrow$   A= 2$^{2009}$-1

$\Rightarrow$B-A = $2^{2009}-\left(2^{2009}-1\right)=2^{2009}-2^{2009}+1=1$

Đại số lớp 6