Câu hỏi của Vũ Minh Hằng

Question

Chứng minh tổng sau chia hết cho 7

A = $2^1+2^2+2^3+2^4+...+2^{59}+2^{60}$

Phương Trâm · Accepted Answer

Giải:

$A=\text{( }2^1+2^2+2^3\text{)}+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)$

$A=2^1.\left(1+2+2^2\right)+2^4.\left(1+2+2^2\right)+...+2^{58}.\left(1+2+2^2\right)$

$A=2.7+2^4.7+...+2^{58}.7$

$A=7.\left(2+2^4+2^{58}\right)⋮7$

$\Rightarrow A=2^1+2^2+2^3+2^4+....+2^{59}+2^{60}$ chia hết cho $7$

Câu trả lời:

Giải:

$A=\text{( }2^1+2^2+2^3\text{)}+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)$

$A=2^1.\left(1+2+2^2\right)+2^4.\left(1+2+2^2\right)+...+2^{58}.\left(1+2+2^2\right)$

$A=2.7+2^4.7+...+2^{58}.7$

$A=7.\left(2+2^4+2^{58}\right)⋮7$

$\Rightarrow A=2^1+2^2+2^3+2^4+....+2^{59}+2^{60}$ chia hết cho $7$

Lê Gia Bảo · Answer

$\Rightarrow A=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+....+\left(2^{58}+2^{59}+2^{60}\right)$

$\Rightarrow A=2^1\left(1+2+4\right)+2^4\left(1+2+4\right)+...+2^{58}\left(1+2+4\right)$

$\Rightarrow A=2^1.7+2^4.7+...+2^{58}.7$

$\Rightarrow A=7\left(2^1+2^4+...+2^{58}\right)$

$\Rightarrow$A chia hết cho 7 vì tích có chứ thừa số 7