Câu hỏi của BLINK KƯ

Question

Chứng minh rằng: S= 2 + 22 + 23 +   24  +..... + 28   Chia hết cho -6

Thierry Henry · Accepted Answer

$=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^7+2^8\right)$

$=2.\left(1+2\right)+2^3.\left(1+2\right)+...+2^7.\left(1+2\right)$

$=3.\left(2+2^3+...+2^7\right)=3.2.\left(1+2^2+2^3+...+2^6\right)$

$=6.\left(1+2^2+2^3+...+2^6\right)⋮-6$Câu trả lời: $=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^7+2^8\right)$

$=2.\left(1+2\right)+2^3.\left(1+2\right)+...+2^7.\left(1+2\right)$

$=3.\left(2+2^3+...+2^7\right)=3.2.\left(1+2^2+2^3+...+2^6\right)$

$=6.\left(1+2^2+2^3+...+2^6\right)⋮-6$

Luân Đào · Answer

$S=2+2^2+2^3+...+2^8$

$=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^6\left(2+2^2\right)$

$=1\cdot6+2^2\cdot6+...+2^6\cdot6$

$=6\left(1+2^2+...+2^6\right)=-6\cdot\left(-1\right)\left(1+2^2+...+2^6\right)⋮\left(-6\right)$Câu trả lời: $S=2+2^2+2^3+...+2^8$

$=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^6\left(2+2^2\right)$

$=1\cdot6+2^2\cdot6+...+2^6\cdot6$

$=6\left(1+2^2+...+2^6\right)=-6\cdot\left(-1\right)\left(1+2^2+...+2^6\right)⋮\left(-6\right)$

Bài 13: Bội và ước của một số nguyên