a, \(M=5+5^2+5^3+...+5^{100}\)
\(\Rightarrow5M=5^2+5^3+5^4+...+5^{101}\)
\(\Rightarrow5M-M=\left(5^2+5^3+5^4+...+5^{101}\right)-\left(5+5^2+5^3+....+5^{100}\right)\)
\(\Rightarrow4M=5^{101}-5\)
\(\Rightarrow M=\frac{5^{101}-5}{4}\)
Vậy : \(M=\frac{5^{101}-5}{4}\)
b, \(N=5^1+5^2+5^3+...+5^{2010}\)
\(N=\left(5^1+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{2009}+5^{2010}\right)\)
\(N=5\left(1+5\right)+5^3\left(1+5\right)+...+5^{2009}\left(5+1\right)\)
\(N=5.6+5^3.6+...+5^{2009}.6\)
\(N=\left(5+5^3+...+5^{2009}\right).6\)
\(\Rightarrow N⋮6\)
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\(N=5^1+5^2+5^3+5^4+...+5^{2010}\)
\(N=\left(5^1+5^2+5^3\right)+\left(5^4+5^5+5^6\right)+...+\left(5^{2008}+5^{2009}+5^{2010}\right)\)
\(N=5\left(1+30\right)+5^4\left(1+30\right)+...+5^{2008}\left(1+30\right)\)
\(N=5.31+5^4.31+...+5^{2008}.31\)
\(N=\left(5+5^4+...+5^{2008}\right).31\)
\(\Rightarrow N⋮31\)