Câu hỏi của Crazy

Question

CMR: 5+5²+5³+...+5³⁰ chia hết cho 6 và 31

headsot96 · Accepted Answer

Ta có $\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{29}+5^{30}\right)=\left(5+5^2\right)+5^2\left(5+5^2\right)+...+5^{28}\left(5+5^2\right)$

$=\left(5+5^2\right)\left(1+5^2+...+5^{28}\right)=30\left(1+5^2+..+5^{28}\right)⋮6$

Ta có $(5+5^2+5^3)+...+\left(5^{28}+5^{29}+5^{30}\right)=(5+5^2+5^3)+...+5^{27}(5+5^2+5^3)$

$=155+...+5^{27}.155=155\left(1+...+5^{27}\right)⋮31$

Câu trả lời:

Ta có $\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{29}+5^{30}\right)=\left(5+5^2\right)+5^2\left(5+5^2\right)+...+5^{28}\left(5+5^2\right)$

$=\left(5+5^2\right)\left(1+5^2+...+5^{28}\right)=30\left(1+5^2+..+5^{28}\right)⋮6$

Ta có $(5+5^2+5^3)+...+\left(5^{28}+5^{29}+5^{30}\right)=(5+5^2+5^3)+...+5^{27}(5+5^2+5^3)$

$=155+...+5^{27}.155=155\left(1+...+5^{27}\right)⋮31$

★๖ۣۜßảo๖ۣۜPɦα♏๖ۣۜ[EηgĻïšħ☯€lub]★ · Answer

-$5+5^2+5^3+...+5^{30}=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{29}+5^{30}\right)$

$=30+5^2\cdot30+5^4\cdot30+...+5^{28}\cdot30=30\left(1+5^2+...+^{28}\right)⋮6$

-$5+5^2+5^3+...+5^{30}=5\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+...+5^{28}\left(1+5+5^2\right)$

$5\cdot31+5^4\cdot31+...+5^{28}\cdot31=31\left(5+5^4+...+5^{28}\right)⋮31$

Câu trả lời:

-$5+5^2+5^3+...+5^{30}=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{29}+5^{30}\right)$

$=30+5^2\cdot30+5^4\cdot30+...+5^{28}\cdot30=30\left(1+5^2+...+^{28}\right)⋮6$

-$5+5^2+5^3+...+5^{30}=5\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+...+5^{28}\left(1+5+5^2\right)$

$5\cdot31+5^4\cdot31+...+5^{28}\cdot31=31\left(5+5^4+...+5^{28}\right)⋮31$