Question

Cho : A = 2¹+ 2²+ 2³+ 2⁴+ 2⁵+ …+ 2⁹⁰ Chứng tỏ A chia hết cho 7

Answer 1

\(A=2^1+2^2+2^3+...+2^{90}\)​

\(=\left(2^1+2^2+2^3\right)+...+\left(2^{88}+2^{89}+2^{90}\right)\)

\(=2^1\left(1+2+2^2\right)+...+2^{88}\left(1+2+2^2\right)\)

​ \(=2^1\cdot7+...+2^{88}\cdot7\)

\(=7\left(2^1+...+2^{88}\right)⋮7\)

Answer 2

A= 2¹+ 2²+ 2³+......+290
= (2¹+2²+2³⁾ +.......+ (2⁸⁸+2⁸⁹+2⁹⁰)
=2¹( 1+2+2²)+ .......+2⁸⁸(1+2+2²)
=2¹.7+......+2⁸⁸.7
=7.(2¹+.....+2⁸⁸_)
=> Tổng A chia hết cho 7