Cho S= 1+2+2^2 + 2^3 + 2^4 + 2^5 +…+ 2^2018 + 2^2019. Chứng tỏ rằng S chia hết cho 3 Cho S= 2+ 2^2 + 2^3 + 2^4 + 2^5 +…

Giải thích các bước giải:

Ta có:
`S=(1+2)+(2^2+2^3)+(2^4+2^5)+…+(2^2018+2^2019)`
`S=3+(2^2 .1+2^2 . 2)+(2^4 .1 +2^4 .2)+…+(2^2018 .1+2^2018 . 2)`
`S=3.1+2^2 .(1+2)+2^4.(1+2)+…+2^2018 .(1+2)`
`S=3.1+2^2 .3+2^4 . 3+…+2^2018 . 3`
`S=3.(1+2^2+2^4+…+2^2018)`
Vì `3\vdots3`
`=>S\vdots3`
Ta có:
`S=2+ 2^2 + 2^3 + 2^4 + 2^5 +…+ 2^12`
`S=(2+2^2+2^3+2^4+2^5+2^6)+(2^7+2^8+2^9+2^10+2^11+2^12)`
`S=126+(2^6 .2+2^6 .2^2+2^6 .2^3 +2^6 .2^4 +2^6 .2^5 +2^6 .2^6)`
`S=126.1+2^6.(2+2^2+2^3+2^4+2^5+2^6)`
`S=126.1+2^6 .126`
`S=126.(1+2^6)`
Vì `126\vdots42`
`=>S\vdots42`