Câu hỏi của Phạm Quỳnh Thư

Question

Chứng minh rằng n = 5 ^ 0 + 5 ^ 1 + + 5 ^ 2019 chia hết cho 126 ;chia hết cho 30

bai nay co hai y lan nha

•ℳเйʑ❤๓ê❤ǥą๓ë⁀ᶜᵘᵗᵉ · Accepted Answer

$n=5^0+5^1+...+5^{2019}$

$n=\left(5^0+5^3\right)+\left(5^1+5^4\right)+...+\left(5^{2016}+5^{2019}\right)$

$n=\left(5^0+5^3\right)+5\left(5^0+5^3\right)+...+5^{2016}\left(5^0+5^3\right)$

$n=126+5\cdot126+...+5^{2016}\cdot126$

$n=126\left(1+5+...+5^{2016}\right)⋮126$ (đpcm)

________

$n=5^0+5^1+...+5^{2019}$

$n=5^0+\left(5^1+5^2\right)+...+\left(5^{2017}+5^{2018}\right)+5^{2019}$

$n=5^0+\left(5^1+5^2\right)+...+5^{2016}\left(5^1+5^2\right)+5^{2019}$

$n=5^0+30+...+5^{2016}\cdot30+5^{2019}$

$n=5^0+30\left(1+5^2+...+5^{2016}\right)+5^{2019}$

Đến đây bí =))

Cbht

Câu trả lời:

$n=5^0+5^1+...+5^{2019}$

$n=\left(5^0+5^3\right)+\left(5^1+5^4\right)+...+\left(5^{2016}+5^{2019}\right)$

$n=\left(5^0+5^3\right)+5\left(5^0+5^3\right)+...+5^{2016}\left(5^0+5^3\right)$

$n=126+5\cdot126+...+5^{2016}\cdot126$

$n=126\left(1+5+...+5^{2016}\right)⋮126$ (đpcm)

________

$n=5^0+5^1+...+5^{2019}$

$n=5^0+\left(5^1+5^2\right)+...+\left(5^{2017}+5^{2018}\right)+5^{2019}$

$n=5^0+\left(5^1+5^2\right)+...+5^{2016}\left(5^1+5^2\right)+5^{2019}$

$n=5^0+30+...+5^{2016}\cdot30+5^{2019}$

$n=5^0+30\left(1+5^2+...+5^{2016}\right)+5^{2019}$

Đến đây bí =))

Cbht