Câu hỏi của Hồ Phúc

Question

Chứng tỏ rằng: 2+2^2+2^3+...+2^2008 chia hết cho 15

Nguyễn Diệu Thu · Accepted Answer

❤1+1=?✿

Câu trả lời:

❤1+1=?✿

Nguyễn Lê Phước Thịnh · Answer

Ta có: $2+2^2+2^3+...+2^{2008}$

$=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{2005}+2^{2006}+2^{2007}+2^{2008}\right)$

$=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{2005}\left(1+2+2^2+2^3\right)$

$=15\left(2+2^5+...+2^{2005}\right)⋮15$

Câu trả lời:

Ta có: $2+2^2+2^3+...+2^{2008}$

$=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{2005}+2^{2006}+2^{2007}+2^{2008}\right)$

$=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{2005}\left(1+2+2^2+2^3\right)$

$=15\left(2+2^5+...+2^{2005}\right)⋮15$