Question

Cho A=3+3²+3³+........+3⁹⁹

Không tính tổng A, hãy chứng tỏ rằngA chia hết cho 13

Answer 1

\(A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)....+\left(3^{97}+3^{98}+3^{99}\right)\)

\(A=3.\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)...+3^{97}.\left(1+3+3^2\right)\)

\(A=3.13+3^4.13+...+3^{97}.13\)

\(A=13.\left(3+3^4+..+3^{97}\right)⋮13\)

Vậy...

Answer 2

\(A=3+3^2+3^3+...+3^{99}\)

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

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

\(A=3\cdot13+...+3^{97}\cdot13\)

\(A=13\cdot\left(3+...+3^{97}\right)⋮13\left(đpcm\right)\)

Answer 3

\(A=3+3^2+3^3+......+3^{99}\)

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

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

\(\Rightarrow A=3.13+.......+3^{97}.13\)

\(\Rightarrow A=13\left(3+....+3^{97}\right)⋮13\left(đpcm\right)\)

Answer 4

ta có: A = 3 + 3² + 3³ + ...+ 3⁹⁹ ( có 99 số hạng)

A = (3+3² + 3³) + ...+ (3⁹⁷ + 3⁹⁸ + 3⁹⁹)   ( có 33 nhóm số hạng)

A = 3.(1+3+3²) + ...+ 3⁹⁷.(1+3+3²)

A = 3.13 + ....+ 3⁹⁷.13

A = 13.(3+...+3⁹⁷) chia hết cho 13

Answer 5

\(A=3+3^2+3^3+...+3^{99}\)

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

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

\(A=3.13+...+3^{97}.13\)

\(A=13.\left(3+...+3^{97}\right)\)

\(\Rightarrow A⋮3\left(đpcm\right)\)