Question

Cho biểu thức:

H= \(3+3^2+3^3+...\) 3^600

Chứng tỏ:

H chia hết cho 13.

H chia hết cho 40.

H chia hết cho 12.

( giúp mk nha mấy bn )

Answer 1

+) Ta có: \(H=3+3^2+3^3+...+3^{600}\)

\(\Rightarrow H=\left(3+3^2+3^3\right)+...+\left(3^{598}+3^{599}+3^{600}\right)\)

\(\Rightarrow H=\left(3+9+27\right)+...+3^{597}.\left(3+3^2+3^3\right)\)

\(\Rightarrow H=39+...+3^{597}.39\)

\(\Rightarrow H=\left(1+...+3^{597}\right).39⋮13\)

\(\Rightarrow H⋮13\)

+) Ta có: \(H=3+3^2+3^3+...+3^{600}\)

\(\Rightarrow H=\left(3+3^2+3^3+3^4+3^5\right)+...+\left(3^{596}+3^{597}+3^{598}+3^{599}+3^{600}\right)\)

\(\Rightarrow H=3\left(1+3+3^2+3^3+3^4\right)+...+3^{596}\left(1+3+3^2+3^3+3^4\right)\)

\(\Rightarrow H=3.40+...+3^{596}.40\)

\(\Rightarrow H=\left(3+...+5^{596}\right).40⋮40\)

\(\Rightarrow H⋮40\)

+) Ta có: \(H=3+3^2+3^3+...+3^{600}\)

\(\Rightarrow H=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{599}+3^{600}\right)\)

\(\Rightarrow H=\left(3+9\right)+3^2\left(3+9\right)+...+3^{598}\left(3+9\right)\)

\(\Rightarrow H=12+3^2.12+...+3^{598}.12\)

\(\Rightarrow H=\left(1+3^2+...+3^{598}\right).12⋮12\)

\(\Rightarrow H⋮12\)

 

Answer 2

\(H=3+3^2+3^3+...+3^{600}\)

\(H=\left(3+3^2+3^3\right)+...+\left(3^{598}+3^{599}+3^{600}\right)\)

\(H=3.\left(1+3+3^2\right)+...+3^{598}.\left(1+3+3^2\right)\)

\(H=3.13+...+3^{598}.13\)

\(H=13.\left(3+...+3^{598}\right)⋮3\)

Vậy H \(⋮\)3