Đáp án:
Ta có :
`S=1+3^2+3^4+3^6+...+3^{2019}+3^{2020}`
`⇒S=(1+3^2)+(3^4+3^6)+....+(3^{2018}+3^{2020})`
`⇒S=(1+3^2)+3^4.(1+3^2)+....+3^{2018}×(1+3^2)`
`⇒S=(1+3^2).(1+3^4+.....+3^{2018})`
`⇒S=10.(1+3^4+..+3^{2018})vdots5`
Do `10.(1+3^4+..+3^{2018})vdots5`
`=> S vdots 5`
Giải thích các bước giải: