Chứng minh tổng S=4+3^2+3^3+…..+3^2000 chia hết cho 40

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1. $S=4+3^2+3^3+…+3^{2000}$

   $=1+3+3^2+3^3+…+3^{2000}$

   $=(1+3+3^2+3^3)+(3^4+3^5+3^6+3^7)+…+(3^{1997}+3^{1998}+3^{1999}+3^{2000})$

   $=(1+3+3^2+3^3)+3^4(1+3+3^2+3^3)+…+3^{1997}(1+3+3^2+3^3)$

   $=40+3^4·40+…+3^{1997}·40$

   $=40(1+3^4+…+3^{1997})\vdots 40$

   $\to S\vdots 40$

    

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