Question

Bài 1: Chứng minh:

\(1+3+3^2+...+3^{119}\)chia hết cho 13

Answer 1

Hứa r phải làm thôi:

Đặt:

\(A=1+3+3^2+.....+3^{119}\)

\(A=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+.....+\left(3^{117}+3^{118}+3^{119}\right)\)

\(A=1\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+....+3^{117}\left(1+3+3^2\right)\)

\(A=1.13+3^3.13+....+3^{117}.13\)

\(A=\left(1+3^3+....+3^{117}\right).13\)

\(A⋮13\rightarrowđpcm\)

Answer 2

Đặt \(A=1+3+3^2+3^3+...+3^{119}\)

\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{117}+3^{118}+3^{119}\right)\)

\(=13+\left(1.3^3+3.3^3+3^2.3^3\right)+...+\left(1.3^{117}+3.3^{117}+3^2.3^{117}\right)\)

\(=13+3^3\left(1+3+3^2\right)+...+3^{117}\left(1+3+3^2\right)\)

\(=13+3^3.13+...+3^{117}.13\)

\(=13.\left(1+3^3+...+3^{117}\right)⋮13\)

Vậy \(A⋮13\)