Đáp án: + Giải thích các bước giải:

Bài `1:`

`a,`

`2+2^2+2^3+2^4`

`=>``(2+2^2)+(2^3+2^4)`

`=>``2(1+2)+2^3(1+2)`
`=>``(1+2)(2+2^3)`

`=>``3*(2+2^3)`

`=>``2+2^2+2^3+2^4vdots3`
`=>ĐPCM`
`b,`

`3+3^2+3^3+``.``.``.+3^20`
`=>``(3+3^2)+(3^3+3^4)+.``.``.+(3^19+3^20)`

`=>``3(1+3)+3^3(1+3)+.``.``.+3^19(1+3)`

`=>``(1+3)(3+3^3+.``.``.+3^19)`

`=>``4(3+3^3+.``.``.+3^19)`

`=>3+3^2+3^3+``.``.``.+3^20vdots4`
`=>ĐPCM`

`3+3^2+3^3+``.``.``.+3^20`
`=>``(3+3^3)+(3^2+3^4)+.``.``.+(3^18+3^20)`

`=>``3(1+9)+3^2(1+9)+.``.``.+3^18(1+9)`

`=>``(1+9)(3+3^2+.``.``.+3^18)`

`=>``10*(3+3^2+.``.``.+3^18)`

`=>3+3^2+3^3+``.``.``.+3^20vdots10`

`=>ĐPCM`
`c,`

`2+2^2+2^3+.``.``.``+2^102`
`=>``(2+2^2+2^3)+(2^4+2^5+2^6)+.``.``.+(2^100+2^101+2^102)`

`=>``2(1+2+4)+2^4(1+2+4)+.``.``.+2^100(1+2+4)`

`=>``(1+2+4)(2+2^4+.``.``.+2^100)`

`=>``7*(2+2^4+.``.``.+2^100)`

`=>2+2^2+2^3+.``.``.``+2^102vdots7`

`=>ĐPCM`

`2+2^2+2^3+.``.``.``+2^102`
`=>``(2+2^3)+(2^2+2^4)+.``.``.+(2^100+2^102)`

`=>``2(1+4)+2^2(1+4)+.``.``.+2^100(1+4)`

`=>``(1+4)(2+2^2+.``.``.+2^100)`

`=>``5*(2+2^2+.``.``.+2^100)`

`=>``2+2^2+2^3+.``.``.``+2^102vdots5`

`=>ĐPCM`

a)2+2²+2³+2⁴ 

=2*(1+2)+`2^3`*(1+2)

=2*3+`2^3`*3 chia hết cho 3

=> 2+2²+2³+2⁴ chia hết cho 3

b)3+3²+3³+…+`3^20`

=3*(1+3)+`3^3`*(1+3)+...+`3^19`*(1+3)

=3*4+`3^3`*4+...+`3^19`*4 chia hết cho 4

=>3+3²+3³+…+`3^20` chia hết cho 4

3+3²+3³+…+`3^20`

=3*(1+`3^2`)+`3^2`*(1+`3^2`)+...+`3^18`*(1+`3^2`)

=3*10+`3^2`*10+...+`3^18`*10 chia hết cho 10

=>3+3²+3³+…+`3^20` chia hết cho 10

c)2+2²+2³+…+`2^102`

=2*(1+2+`2^2`)+`2^4`*(1+2+`2^2`)+...+`2^100`*(1+2+`2^2`)

=2*7+`2^4`*7+...+`2^100`*7 chia hết cho 7

=>2+2²+2³+…+`2^102` chia hết cho 7

2+2²+2³+…+`2^102`

=2*(1+`2^2`)+`2^2`*(1+`2^2`)+...+`2^100`*(1+`2^2`)

=2*5+`2^2`*5+...+`2^100`*5 chia hết cho 5

=>2+2²+2³+…+`2^102` chia hết cho 5

cho mk hay nhất nha