Ta đặt: \(A=5^1+5^2+...+5^{2010}\\ A=\left(5^1+5^2\right)+...+\left(5^{2009}+5^{2010}\right)\\ A=5\left(1+5\right)+...+5^{2009}\left(1+5\right)\\ A=6\left(5+...+5^{2009}\right)⋮6\\ =>A⋮6->\left(a\right)\\ Ta-lại-có:A=5^1+5^2+...+5^{2010}\\ A=\left(5^1+5^2+5^3\right)+...+\left(5^{2008}+5^{2009}+5^{2010}\right)\\ A=5\left(1+5+25\right)+...+5^{2008}\left(1+5+25\right)\\ A=31\left(5+...+2^{2008}\right)⋮31\\ =>A⋮31->\left(b\right)\\ Từ\left(a\right),\left(b\right)=>A⋮6;A⋮31\)
Ta có: 5^1+5^2+5^3+5^4+...+ 5^2010
= (5^1+5^2)+(5^3+5^4)+...+(5^2009 5^2010)
= 5(1+5)+5^3(1+5)+...+5^2009(1+5)
= 6(5+5^5+...+5^2009) chia hết cho 6.
Ta lại có: 5^1+5^2+5^3+5^4+...+ 5^2010
= (5^1+5^2+5^3)+(5^4+5^5+5^6)+...+(5^2008+5^2009+5^2010)
=5(1+5+25)+5^4(1+5+25)+...+5^2008(1+5+25)
= 31(5+5^4+...+5^2008) chia hết cho 31
\(\left(5+5^2\right)+\left(5^3+5^4\right)+....+\left(5^{2009}+5^{2010}\right)=5\left(1+5\right)+5^3\left(1+5\right)+...+5^{2009}\left(1+5\right)\)
= \(6\left(5+5^5+....+5^{2009}\right)chia\)\(hết\) \(cho\) 6
