Question

Chứng tỏ rằng: \(3^1+3^2+3^3+3^4+...+3^{99}+3^{100}\) chia hết cho 40

Answer 1

= \(3\left(1+3+3^2+3^3\right)+...+3^{97}\left(1+3+3^2+3^3\right)\)

=\(40\left(1+...+3^{97}\right)\) chia hết cho 40 

Answer 2

\(3+3^2+3^3+3^4+...+3^{99}+3^{100}\)

\(=\left(3+3^2+3^3+3^4\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)

\(=3.\left(3+3^2+3^3+3^4\right)+...+3^{97}.\left(3+3^2+3^3+3^4\right)\)

\(=3.120+...+3^{97}.120\)

\(=120.\left(3+...+3^{97}\right)\)chia hết cho 40 (đpcm)