A=1+3+3^2+3^3++…+3^11.Chứng tỏ B chia hết cho 13,chia hết cho 40

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1. `A=1+3+3^2+3^3+3^4+….+3^11`

   `->A=(1+3+3^2)+(3^3+3^4+3^5)+…+(3^9+3^10+3^11)`

   `->A=(1+3+3^2)+3^3(1+3+3^2)+…+3^9(1+3+3^2)`

   `->A=13.1+3^{3}.13+3^{9}.13`

   `->A=13(1+3^3+…+3^9)` 

   Vì `13\vdots13`

   `->13(1+3^3+…+3^9)\vdots13`

   `->A\vdots13`

   `A=1+3+3^2+3^3+3^4+….+3^11`

   `->A=(1+3+3^2+3^3)+(3^4+3^5+3^6+3^7)+(3^8+3^9+3^10+3^11)`

   `->A=(1+3^2+3^3)+3^4(3+3^2+3^3)+3^8(3+3^2+3^3)`

   `->A=40.1+3^{4}.40+3^{8}.40`

   `->A=40(1+3^4+3^8)`

   Vì `40\vdots40`

   `->40.1+3^4.40+3^8.40\vdots`

   `->A\vdots40`

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2. Chứng minh $A$ chia hết cho $13$

   $A = 1+3 + 3^2 + 3^3 + …. + 3^{11}$

   $⇔ A = (1+3+3^2) + (3^3 + 3^4 + 3^5) + …. + (3^9 + 3^{10} + 3^{11})$

   $⇔ A = 13 + 3^3.(1+3+3^2) + …. + 3^9.(1+3^1+3^2)$

   $⇔ A = 13 + 3^3. 13 + …. + 3^9 . 13$

   $⇔ A = 13.(1+3^3 + … + 3^9) \vdots 13$($đ.p.c.m$)

   Chứng minh $A$ chia hết cho $40$

   $A = 1+3 + 3^2 + 3^3 + …. + 3^{11}$

   $⇔ A = (1+3+3^2+3^3) + (3^4 + 3^5 + 3^6+3^7)  + (3^8 + 3^9 + 3^{10} + 3^{11})$

   $⇔ A = 40 + 3^4.(1+3 + 3^2 + 3^3) + 3^8.(1 + 3 + 3^2 + 3^3)$

   $⇔ A =40 + 3^4. 40 + …. + 3^8 . 40$

   $⇔ A = 40.(1+3^4 + … + 3^8) \vdots 40$($đ.p.c.m$).

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