a,chứng minh rằng A=1+2+2²+....+2¹¹⁹ chia hết cho 3,7,31 b, tìm số dư của b=1+5+5²+5³+...+5¹⁰⁰ khi chia cho 6,31

`a)` Ta có : `A=1+2+2^2+2^3+..........+2^118+2^119`

`= (1+2)+(2^2+2^3)+..........+(2^118+2^119)`

`= (1+2)+2^2 .(1+2)+..........+ 2^118.(1+2)`

`= 3+2^2 .3+.........+2^118 .3`

`= 3.(1+2^2+......+2^118) \vdots 3`

Lại có : `A=1+2+2^2+2^3+2^4+2^5..........+2^117+2^118+2^119`

`= (1+2+2^2)+(2^3+2^4+2^5)+............+(2^117+2^118+2^119)`

`= (1+2+2^2)+2^3 .(1+2+2^2)+............+2^117.(1+2+2^2)`

`= 7+2^3 .7+........+2^117.7`

`= 7.(1+2^3+......+2^117) \vdots 7`

Lại có : `A=1+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+.............+2^115+2^116+2^117+2^118+2^119`

`=(1+2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8+2^9)+.............+(2^115+2^116+2^117+2^118+2^119)`

`= (1+2+2^2+2^3+2^4)+2^5. (1+2+2^2+2^3+2^4)+.........+2^115 .(1+2+2^2+2^3+2^4)`

`= 31+2^5 .31+........+2^115.31`

`= 31.(1+2^5+......+2^115) \vdots 31`

Vậy `A \vdots 3;7;31` ( điều phải chứng minh )

`b)` Ta có : `B=1+5+5^2+5^3+.......+5^99+5^100`

`= (1+5)+(5^2+4^3)+........+(5^99+5^100)`

`= (1+5)+5^2 .(1+5)+.......+5^99 .(1+5)`

`= 6+5^2 .6+......+5^99 .6`

`= 6.(1+5^1+.....5^99) \vdots 6`

Lại có : `B=1+5+5^2+5^3+5^4+5^5+.......+5^98+5^99+5^100`

`= (1+5+5^2)+(5^3+5^4+5^5)+.......+(5^98+5^99+5^100)`

`= (1+5+5^2)+5^2 .(1+5+5^2)+..... +5^98 .(1+5+5^2)`

`= 31+5^2 .31+........5^98 .31`

`= 31.(1+5^2+.......+5^98) \vdots 31`

Vậy `B` chia cho `6` và `31` dư `0`