Cho A = 1+3+3^2 + 3^3 +… + 3^ 1999 +3^2000 . Chứng minh rằng A chia hết cho 3

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1. Đáp án:

   `A\vdots13`

   Giải thích các bước giải:

   `text{Cho}`
   `A = 1+3+3^2 + 3^3 +… + 3^ 1999 +3^2000`
   `text{Chứng minh:}` `A\vdots13 .`
   `text{Ta có:}`
   `A=(1+3+3^2)+(3^3+3^4+3^5)+(3^6+3^7+3^8)+…+(3^1998+3^1999+3^2000)`
   `A=(1+3+9)+(3^3+3^4+3^5)+(3^6+3^7+3^8)+…+(3^1998+3^1999+3^2000)`
   `A=(4+9)+(3^3+3^4+3^5)+(3^6+3^7+3^8)+…+(3^1998+3^1999+3^2000)`
   `A=13+(3^3 .1+3^3 . 3+3^3 .3^2)+(3^6 . 1 +3^6 .3+3^6 . 3^2)+…+(3^1998 . 1 +3^1998 . 3 +3^1998 . 3^2)`
   `A=13.1+3^3.(1+3+3^2)+3^6.(1+3+3^2)+…+3^1998.(1+3+3^2)`
   `A=13.1+3^3.(1+3+9)+3^6.(1+3+9)+…+3^1998.(1+3+9)`
   `A=13.1+3^3.(4+9)+3^6.(4+9)+…+3^1998.(4+9)`
   `A=13.1+3^3 . 13+3^6 . 13+…+3^1998 . 13`
   `A=13.(1+3^3+3^6+…+3^1998)`
   `text{Vì}` `13\vdots13`
   `=>13.(1+3^3+3^6+…+3^1998)\vdots13`
   `text{Vậy}` `A\vdots13`

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