Question

Chứng minh 1+2 + 2^2 +...+ 2^98 chia hết cho 7

Answer 1

\(1+2+2^2+...+2^{98}\\ =\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+...+\left(2^{96}+2^{97}+2^{98}\right)\\ =1\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+...+2^{96}\left(1+2+2^2\right)\\ =1.7+2^3.7+...+2^{96}.7\\ =7\left(1+2^3+...+2^{96}\right)⋮7\left(dpcm\right)\)

Answer 2

\(1+2+2^2+.....+2^{98}\)

\(=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+...+\left(2^{96}+2^{97}+2^{98}\right)\)

\(=1\left(1+2+2^2\right)+2^3\left(1+2+2^3\right)+....+2^{96}\left(1+2+2^2\right)\)

\(=1.7+2^3.7+...+2^{96}.7\)

\(=7\left(1+2^3+....+2^{96}\right)⋮7\left(đpcm\right)\)

Chúc bạn học tốt!