Question

Ai làm được mik tích cho, hạn là ngày hôm nay (Cứu!!!!!!)

Answer 1

Bài 1: `m = 3 + 3^2 + ... + 3^2019`.

`= 3(1+3+3^2) + 3^4(1+3+3^2) + 3^7(1+3+3^2) + ... + 3^2017(1+3+3^2)`

`= 13 . 3 + 13 . 3^4 + 3^7 . 13 + ... + 3^2017 .13`

`= 13(3+3^4+3^7+...+3^2017) vdots 13`.

Answer 2

Bài 4: `(n+2021^2022)(n+2022^2021)`

`@ n cancel vdots 2 => n + 2021^2022 equiv 1 + 1 equiv 2 equiv 0 ( mod 2)`

`@ n vdots 2 <=> n + 2022^2021 equiv 0 + 0 equiv 0 ( mod 2)`

Vậy ta có dpcm

Answer 3

3:

\(A=\left(5+5^2+5^3+5^4\right)+5^4\left(5+5^2+5^3+5^4\right)+...+5^{2020}\left(5+5^2+5^3+5^4\right)\)

=780(1+5^4+...+5^2020) chia hết cho 65

2:

\(A=3\left(1+3^2+3^4\right)+...+3^{2011}\left(1+3^2+3^4\right)\)

=91(3+...+3^2011) chia hết cho 13

A=3(1+3^2+3^4+3^6)+...+3^2009(1+3^2+3^4+3^6)

=820*(3+...+3^2009) chia hết cho 41