Từ \(2^1\rightarrow2^{100}\) có 100 số.
\(\circledast\) Cứ 4 số thành 1 nhóm, ta được: \(100:4=25\)(nhóm)
Viết: \(H=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(\Leftrightarrow H=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{97}\left(1+2+2^2+2^3\right)\)
\(\Leftrightarrow H=2.15+2^5.15+...+2^{97}.15\)
\(\Leftrightarrow H=15\left(2+2^5+...+2^{97}\right)\)
Vì \(15\) \(⋮\) \(5\) \(\Rightarrow15\left(2+2^5+...+2^{97}\right)⋮5\)
\(\Rightarrow H\) \(⋮\) \(5\) (1)
\(\circledast\) Cứ 5 số thành 1 nhóm, ta được: \(100:5=20\)(nhóm)
Viết: \(H=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(\Leftrightarrow H=2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)\)
\(\Leftrightarrow H=2.31+2^6.31+...+2^{96}.31\)
\(\Leftrightarrow H=31\left(2+2^6+...+2^{96}\right)\)
Vì \(31\) \(⋮\) \(31\) \(\Rightarrow31\left(2+2^6+...+2^{96}\right)⋮31\)
\(\Rightarrow H\) \(⋮\) \(31\) (2)
Từ (1) và (2) \(\Rightarrow H\) \(⋮\) \(155.\)
