a)\(C=2+2^2+2^3+....+2^{100}\)
\(=\left(2+2^2+2^3+2^4+2^5\right)+2^5\left(2+2^2+2^3+2^4+2^5\right)+...+2^{95}\left(2+2^2+2^3+2^4+2^5\right)\)
\(=62+2^5.62+...+2^{95}.62=62\left(1+2^5+...+2^{95}\right)=31.2\left(1+2^5+....+2^{95}\right)⋮31\)
\(\Rightarrow C⋮31\)
=>đccm
\(C=2+2^2+2^3+...+2^{99}+2^{100}\)
\(C=\left(2+2^2+2^3+2^4+2^5\right)+....+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(C=2.\left(1+2+2^2+2^3+2^4\right)+....+2^{96}.\left(1+2+2^2+2^3+2^4\right)\)
\(C=31.2+.....+2^{96}.31=31.\left(2+....+2^{96}\right)⋮31\)
Suy ra \(C⋮31\)
b) Ta có \(2.C=2^2+2^3+2^4+....+2^{99}+2^{100}+2^{101}\)
Suy ra \(2.C-C=2^{101}-2\)hay \(C=2^{101}-2\)
Khi đó \(2^{2x-1}-2=2^{101}-2\)
\(\Rightarrow2^{2x-1}=2^{101}\)
\(\Rightarrow2x-1=101\Rightarrow2x=100\Rightarrow x=50\)
Vậy x = 50