Toán Chứng minh tổng S=4+3^2+3^3+…..+3^2000 chia hết cho 40 30/11/2021 By Ariana Chứng minh tổng S=4+3^2+3^3+…..+3^2000 chia hết cho 40
$S=4+3^2+3^3+…+3^{2000}$ $=1+3+3^2+3^3+…+3^{2000}$ $=(1+3+3^2+3^3)+(3^4+3^5+3^6+3^7)+…+(3^{1997}+3^{1998}+3^{1999}+3^{2000})$ $=(1+3+3^2+3^3)+3^4(1+3+3^2+3^3)+…+3^{1997}(1+3+3^2+3^3)$ $=40+3^4·40+…+3^{1997}·40$ $=40(1+3^4+…+3^{1997})\vdots 40$ $\to S\vdots 40$ Trả lời
$S=4+3^2+3^3+…+3^{2000}$
$=1+3+3^2+3^3+…+3^{2000}$
$=(1+3+3^2+3^3)+(3^4+3^5+3^6+3^7)+…+(3^{1997}+3^{1998}+3^{1999}+3^{2000})$
$=(1+3+3^2+3^3)+3^4(1+3+3^2+3^3)+…+3^{1997}(1+3+3^2+3^3)$
$=40+3^4·40+…+3^{1997}·40$
$=40(1+3^4+…+3^{1997})\vdots 40$
$\to S\vdots 40$