Question

[Phùng Nhật Trường]

Cho A=2² +2⁴ +2⁶ +2⁸ +...+2¹⁸ +2²⁰

^{Chứng tỏ rằng A chia hết cho 5}

Answer 1

[Lê Minh Tú]

\(A=2^2+2^4+2^6+2^8+...+2^{18}+2^{20}\)

\(=\left(2^2+2^4\right)+\left(2^6+2^8\right)+...+\left(2^{18}+2^{20}\right)\)

\(=2^2.\left(1+2^2\right)+2^6.\left(1+2^2\right)+...+2^{18}.\left(1+2^2\right)\)

\(=2^2.5+2^6.5+...+2^{18}.5\)

\(=5.\left(2^2+2^6+...+2^{18}\right)⋮5\)(vì trong tích có 1 thừa số là 5)

\(\Rightarrow A⋮5\)

Answer 2

[Edogawa Conan]

\(A=2^2+2^4+2^6+2^8+.....+2^{20}\)

\(A=\left(2^2+2^4\right)+\left(2^6+2^8\right)+.....+\left(2^{18}+2^{20}\right)\)

\(A=2^2\left(1+2^2\right)+2^6\left(1+2^2\right)+......+2^{18}\left(1+2^2\right)\)

\(A=2^2.5+2^6.5+......+2^{18}.5\)

\(A=5\left(2^2+2^6+.....+2^{18}\right)\)

vì \(5⋮5\Rightarrow5\left(2^2+2^6+.....+2^{18}\right)⋮5\)

\(\Rightarrow A⋮5\)

chúc bạn học giỏi ^^