Question

1 cho tong M = 2 + 2^2 + 2^3 + ... + 2^120

chứng minh M ⋮ 105

Answer 1

ta có : \(M=2+2^2+2^3+2^4+...+2^{120}\)

\(\Leftrightarrow M=\left(2+2^2+...+2^{12}\right)+\left(2^{13}+2^{14}+...+2^{24}\right)+...+\left(2^{109}+2^{110}+...+2^{120}\right)\)

\(\Leftrightarrow M=2\left(1+2+...+2^{11}\right)+2^{13}\left(1+2+...+2^{11}\right)+...+2^{109}\left(1+2+...+2^{11}\right)\)

\(\Leftrightarrow M=\left(2+2^{13}+...+2^{109}\right)\left(1+2+...+2^{11}\right)\)

\(\Leftrightarrow M=4095\left(2+2^{13}+...+2^{109}\right)=105.39\left(2+2^{13}+...+2^{109}\right)⋮105\left(đpcm\right)\)