Question

Cho A = 3 + 3² + 3³ +...+ 3⁹ + 3¹⁰

chứng minh rằng A chia hết cho 4

Answer 1

\(A=3+3^2+...+3^9+3^{10}\)

\(=\left(3+3^2\right)+...+\left(3^9+3^{10}\right)\)

\(=3\left(1+3\right)+...+3^9\left(1+3\right)\)

\(=3\cdot4+...+3^9\cdot4\)

\(=4\left(3+...+3^9\right)⋮4\)

Answer 2

Ta có:

\(A=3+3^2+...+3^{10}\)

\(\Rightarrow A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^9+3^{10}\right)\)

\(\Rightarrow A=3\left(1+3\right)+3^3\left(1+3\right)+...+3^9\left(1+3\right)\)

\(\Rightarrow A=3.4+3^3.4+...+3^9.4\)

\(\Rightarrow A=\left(3+3^3+...+3^9\right).4⋮4\)

\(\Rightarrow A⋮4\)

Vậy \(A⋮4\)