Question

chứng tỏ tổng sau : A= 2^2+2^4+2^6+2^8+...+2^18+2^20  chia hết cho 5

Giúp mình với, làm ơn ạ !! Mình cần gấpp :(

Answer 1

Ta có : A=2²+2⁴+2⁶+...+2²⁰

=(2²+2⁴)+(2⁶+2⁸)+...+(2¹⁸+2²⁰)

=2²(1+2²)+2⁶(1+2²)+...+2¹⁸(1+2²)

=2².5+2⁶.5+...+2¹⁸.5 chia hết cho 5

Vậy A chia hết cho 5.

Answer 2

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

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

\(\Leftrightarrow A=20+2^4.\left(2^2+2^4\right)+...+2^{16}.\left(2^2+2^4\right)\)

\(\Leftrightarrow A=20+2^4.20+..+2^{16}.20\)

\(\Leftrightarrow A=20\left(1+2^4+..+2^{16}\right)\)

Vì \(20⋮5\)

\(\Rightarrow A=20\left(1+2^4+..+2^{16}\right)⋮5\)

Vậy \(A⋮5\)

hok tốt!!

Answer 3

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

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

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

\(A=2^2\cdot5+2^6\cdot5+...+2^{18}\cdot5\)

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

=> A chia hết cho 5

Answer 4

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

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

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

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

Vậy \(A\)chia hết cho \(5\)