Question

 

chứng minh A=2^2+2^4+2^6+...+2^18+2^20 chia hết cho 5

Answer 1

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

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

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

\(=2^2.5+2^6.5+...+2^{18}.5⋮5\)

\(\Rightarrow A⋮5\)

Answer 2

Dễ dàng tính được \(A=\frac{2^{22}-2^2}{3}\)

Ta có:

\(A=\frac{2^2\cdot\left(2^{20}-1\right)}{3}=\frac{2^2\cdot\left(2^{10}-1\right)\cdot\left(2^{10}+1\right)}{3}=\frac{2^2\cdot1023\cdot1025}{3}\)

Do đó A chia hết cho 5

Chúc bạn học tốt!