Question

A=1+2+2²+...+2²⁰²⁰+2²⁰²¹ chứng minh Achia hết cho 7 

Answer 1

giúp với

 

Answer 2

\(A=1+2+2^2+...+2^{2020}+2^{2021}\\ A=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+...+\left(2^{2019}+2^{2020}+2^{2021}\right)\\ A=7+2^3.7+...+2^{2019}.7\\ A=7.\left(1+2^3+...+2^{2012}\right)⋮7\)

Answer 3

A=(1+2+2^2)+............+(2^2019+2^2020+2^2021)

=1.(1+2+2^2)+.........+2^2019.(1+2+2^2)

1.7+.............+2^2019.7

7.(1+...........+2^2019)

vì 7chia hết cho 7

  Vậy A chia hết cho 7