Question

Cho A = 3 + 3^2 + 3^3 + 3^4 + ... + 3^50

Chứng minh A chia hết cho 4

Ai làm nhanh nhất mk sẽ k 

Answer 1

\(A=3+3^2+3^3+3^4+...+3^{50}.\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{49}+3^{50}\right)\)

\(=\left(3\cdot1+3\cdot3\right)+\left(3^3\cdot1+3^3\cdot3\right)+...+\left(3^{49}\cdot1+3^{49}\cdot3\right)\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{49}\left(1+3\right)\)

\(=3\cdot4+3^3\cdot4+...+3^{49}\cdot4\)

\(=4\cdot\left(3+3^3+...+3^{49}\right)⋮4\)

\(\Rightarrow A⋮4\)

Học tốt ^3^

Answer 2

Trả lời:

\(A=3+3^2+3^3+3^4+...+3^{50}\)

\(A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{49}+3^{50}\right)\)

\(A=3.\left(1+3\right)+3^3.\left(1+3\right)+...+3^{49}.\left(1+3\right)\)

\(A=\left(3+3^3+...+3^{49}\right).4\)

Vì \(3+3^3+...+3^{49}\inℕ\)

Mà \(4⋮4\)

\(\Rightarrow\)\(\left(3+3^3+...+3^{49}\right).4⋮4\)

Hay \(A⋮4\left(đpcm\right)\)

Vậy\(A⋮4\)

Hok tốt!

Vuong Dong Yet