Question

ChoA=2+2²+2³+...+2²⁰

CMR A :. 5

( chứng minh rằng A chia hết cho 5)

Answer 1

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

\(A=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\)

\(A=\left(2.1+2.2+2.2^2+2.2^3\right)+\left(2^5.1+2^5.2+2^5.2^2+2^5.2^3\right)+...\left(2^{17}.1+2^{17}.2+2^{17}.2^2+2^{17}.2^3\right)\)

\(A=2.\left(1+2+4+8\right)+2^5.\left(1+2+4+8\right)+...+2^{17}.\left(1+2+4+8\right)\)

\(A=2.15+2^5.15+...+2^{17}.15\)

\(A=15.\left(2+2^5+...+2^{17}\right)\)

Vì 15 chia hết cho 5 

=> A chia hết cho 5

Answer 2

A=2.(1+2+4+8)+...2^17(1+2+4+8)

A=2.15+2^5.15+...+2^17.15

A=15.(2+2^5+...+2^17) chia het cho 5

Vay.............