Câu hỏi của LÊ TRẦN BÁCH

Question

A=21+22+23+...+22016

chứng tỏ A chia hết cho 7

Nguyễn Đức Trí · Accepted Answer

$A=2^1+2^2+2^3+...+2^{2016}$

$\Rightarrow A=2\left(1+2^1+2^2\right)+2^4\left(1+2^1+2^2\right)...+2^{2014}\left(1+2^1+2^2\right)$

$\Rightarrow A=2.7+2^4.7...+2^{2014}.7$

$\Rightarrow A=7\left(2+2^4...+2^{2014}\right)⋮7$

$\Rightarrow dpcm$Câu trả lời: $A=2^1+2^2+2^3+...+2^{2016}$

$\Rightarrow A=2\left(1+2^1+2^2\right)+2^4\left(1+2^1+2^2\right)...+2^{2014}\left(1+2^1+2^2\right)$

$\Rightarrow A=2.7+2^4.7...+2^{2014}.7$

$\Rightarrow A=7\left(2+2^4...+2^{2014}\right)⋮7$

$\Rightarrow dpcm$