Giải giúp mình : CMR:(3^102_1)chia hết cho 8 Tính tổng:A=1_2+3_4+...+2019_2020+2021

Giải thích các bước giải:

a.Ta có:

$3^{102}-1=(3^2)^{51}-1=9^{51}-1\quad\vdots\quad 9-1=8$

$\to đpcm$

b.Ta có:

$A=1.2+3.4+...+2019.2020+2021$

Ta có:

$B=1.2+3.4+...+2019.2020$

$\to B=1.(1+1)+3.(3+1)+...+2019.(2019+1)$

$\to B=1^2+1+3^2+3+...+2019^2+2019$

$\to B=1^2+3^2+...+2019^2+(1+3+..+2019)$

$\to B=(1^2+3^2+...+2019^2)+(2^2+4^2+...+2018^2)-(2^2+4^2+...+2018^2)+(1+3+..+2019)$

$\to B=(1^2+2^2+...+2018^2+2019^2)-4(1^2+2^2+...+1009^2)+(1+3+...+2019)$

Áp dụng công thức $1^2+2^2+...+n^2=\dfrac{n(n+1)(2n+1)}{6}$

$\to B=\dfrac{2019(2019+1)(2\cdot 2019+1)}{6}-4\cdot \dfrac{1009(1009+1)(2\cdot 1009+1)}{6}+\dfrac{(2019+1)\cdot 1010}{2}$

$\to B=1374754430$