Question

Chung minh tong s=1+2+2^2+2^3+.....+2^59 chia het cho 3

Answer 1

Ta có S=1+2+2²+2³+...+2⁵⁹

\(\Rightarrow\)2S=2+2²+2³+2⁴+...+2⁶⁰

\(\Rightarrow\)2S-S=2⁶⁰-1

do 2 chia 3 dư 1 \(\Rightarrow\)2⁶⁰ chia 3 dư 1⁶⁰\(\Rightarrow\)2⁶⁰ chia 3 dư 1

\(\Rightarrow\)2⁶⁰ -1 \(⋮\)3

Hay S\(⋮\)3 (dpcm)

Answer 2

\(1+2+2^2+2^3+...+2^{59}\)

\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{58}+2^{59}\right)\)

\(=3+2^2\left(1+2\right)+...+2^{58}\left(1+2\right)\)

\(=3+2^2\times3+...+2^{58}\times3\)

\(=3\times\left(1+2^2+...+2^{58}\right)⋮3\)

Vậy \(S⋮3\)