Question

Chứng tỏ rằng H chia hết ho 115, biết H=2+2^2+2^3+2^4+..........+2^97+2^98+2^99+2^100 .

Nguyễn Thế Bảo, help me !

Answer 1

\(H=2+2^2+...+2^{100}\)

\(\Rightarrow H=\left(2+2^2+2^3+2^4\right)+...+\left(2^{97}+2^{98}+2^{99}+2^{100}\right)\)

\(\Rightarrow H=2\left(1+2+2^2+2^3\right)+...+2^{97}\left(1+2+2^2+2^3\right)\)

\(\Rightarrow H=2.15+...+2^{97}.15\)

\(\Rightarrow H=\left(2+...+2^{97}\right).15⋮15\)

\(\Rightarrow H⋮15\left(đpcm\right)\)