Question

Chứng minh tổng A=2+2²⁺2^3+...+2¹¹⁸+2¹¹⁹+2¹²⁰ chia hết cho 7

Answer 1

\(A=\left(2+2^2+2^3\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\\ A=2\left(1+2^2+2^3\right)+...+2^{118}\left(1+2^2+2^3\right)\\ A=\left(1+2^2+2^3\right)\left(2+...+2^{118}\right)\\ A=7\left(2+...+2^{118}\right)⋮7\)

Answer 2

\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{118}\left(1+2+2^2\right)\)

\(=2.7+2^4.7+...+2^{118}.7=7\left(2+2^4+...+2^{118}\right)⋮7\)