Question

cmr 2 2^2 2^3 ... 2^60 chia hết cho 3,5,7

Answer 1

Đặt: \(S=2+2^2+2^3+...+2^{60}\)

\(S=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\cdot\left(1+2\right)\)

\(S=3\cdot\left(2+2^3+...+5^{59}\right)\)

Vậy S chia hết cho 3

________________________

\(S=2+2^2+2^3+...+2^{60}\)

\(S=\left(2+2^3\right)+\left(2^2+2^4\right)+...+\left(2^{58}+2^{60}\right)\)

\(S=2\left(1+2^2\right)+2^2\left(1+2^2\right)+...+2^{58}\left(1+2^2\right)\)

\(S=5\left(2+2^2+....+2^{58}\right)\)

Vậy S chia hết cho 5 

___________________________

\(S=2+2^2+2^3+...+2^{60}\)

\(S=2\left(1+2+2^2\right)+2^2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

\(S=7\cdot\left(2+2^2+...+2^{58}\right)\)

Vậy  S chia hết cho 7