Câu hỏi của Trần Phước Nghĩa

Question

chứng tỏ Rằng A= 2+2²+2³+...+2¹⁰⁰chia hết cho 6

ILoveMath · Accepted Answer

$A=2+2^2+2^3+...+2^{100}$

$\Rightarrow A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)$

$\Rightarrow A=6+2^3\left(2+2^2\right)+...+2^{99}\left(2+2^2\right)$

$\Rightarrow A=6+2^3.6+...+2^{99}.6$

$\Rightarrow A=6\left(1+2^3+...+2^{99}\right)⋮6$

Câu trả lời:

$A=2+2^2+2^3+...+2^{100}$

$\Rightarrow A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)$

$\Rightarrow A=6+2^3\left(2+2^2\right)+...+2^{99}\left(2+2^2\right)$

$\Rightarrow A=6+2^3.6+...+2^{99}.6$

$\Rightarrow A=6\left(1+2^3+...+2^{99}\right)⋮6$