Question

Tính tổng :A=3+3^2+3^3+3^4+....+3^20 chia hết cho 4 ,cho 40

Giải giúp mình nhé! Thank you! Minh sẽ cho 5tick

Answer 1

\(A=3+3^2+3^3+...+3^{20}\)

\(A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{19}+3^{20}\right)\)

\(A=3\left(1+3\right)+3^3\left(3+1\right)+...+3^{19}\left(1+3\right)\)

\(\Rightarrow A=4\left(3+3^3+...+3^{19}\right)\)

\(\Rightarrow A⋮4\)

Answer 2

\(A=3+3^2+3^3+...+3^{20}\)

\(\Rightarrow3A=3^2+3^3+...+3^{20}+3^{21}\)

\(\Rightarrow3A-A=\left(3^2+3^3+...+3^{21}\right)-\left(3+3^2+....+3^{20}\right)\)

\(\Rightarrow2A=3^{21}-3\)

\(\Rightarrow A=\frac{3^{21}-3}{2}\)

Answer 3

Còn chứng tỏ thì làm sao bạn