Question

Cho tổng A=1+2+2^2+2^3+...+2^99

CMR: A không chia hết cho 7

Answer 1

\(A=1+2+2^2+2^3+...+2^{99}\)

\(=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+...+\left(2^{96}+2^{97}+2^{98}\right)+2^{99}\)

\(=7+2^3\left(1+2+2^2\right)+...+2^{96}\left(1+2+2^2\right)+2^{99}\)

\(=7+2^3.7+...+2^{96}.7+2^{99}\)

\(=7\left(1+2^3+...+2^{96}\right)+2^{99}\)

Vì \(7⋮7=>7\left(1+2^3+...+2^{96}\right)⋮7\) mà \(2^{99}⋮̸7\)

\(=>A⋮̸7\)