cho A=1+3+3^2+3^3+3^4+…+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`

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

   `=(1+3+3^2)+3^3(1+3+3^2)+…+3^9(1+3+3^2)`

   `=13.1+3^{3}.13+3^{9}.13`

   `=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`

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

   `=(1+3^2+3^3)+3^4(3+3^2+3^3)+3^8(3+3^2+3^3)`

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

   `=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. $A = 1 + 3 + 3^2 + 3^3 + 3^4 +\dots + 3^{11}$

   $\to A = (1+3+3^2) + (3^3 + 3^4+3^5)+\dots + (3^9 + 3^{10} + 3^{11})$

   $\to A = (1 + 3 + 3^2)+ 3^3(1 + 3 + 3^2) + \dots + 3^9(1 + 3 + 3^2)$

   $\to A = (1 + 3 + 3^2)(1 + 3^3 +\dots +3^9)$

   $\to A = 13.(1 + 3^3 +\dots +3^9)$

   $\to A \quad \vdots \quad 13$

   $A = 1 + 3 + 3^2 + 3^3 + 3^4 +\dots + 3^{11}$

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

   $\to A = (1+3+3^2+ 3^3) + 3^4(1+3+3^2+ 3^3) + 3^8(1+3+3^2+ 3^3)$

   $\to A = (1+3+3^2+ 3^3)(1 + 3^4 + 3^8)$

   $\to A = 40(1 + 3^4 + 3^8)$

   $\to A \quad \vdots \quad 40$

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