cho a=5+5^2+5^3+5^4+…+5^120 . chứng minh rằng a chia hết cho 186

0 bình luận về “cho a=5+5^2+5^3+5^4+…+5^120 . chứng minh rằng a chia hết cho 186”

1. $A=5+5^2+5^3+5^4+…+5^{120}$

   $=(5+5^2)+(5^3+5^4)+…+(5^{119}+5^{120})$

   $=5.(1+5)+5^3.(1+5)+…+5^{119}.(1+5)$

   $=6.(5+5^3+…+5^{119})$ ⋮ $6$

   $⇒A$ ⋮ $6$

   $A=5+5^2+5^3+5^4+…+5^{120}$

   $=(5+5^2+5^3)+…+(5^{118}+5^{119}+5^{200})$

   $=5.(1+5+5^2)+…+5^{118}.(1+5+5^2)$

   $=5.31+…+5^{118}.31$

   $=31.(5+…+5^{118})$ ⋮ $31$

   $⇒A$ ⋮ $31$

   Mà $(6;31)=1$

   $⇒A$ ⋮ $186$.

    

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2. Ta phân tích : `186=31.6`

   Vì `ƯCLN(31;6) ⇒`Ta cần chứng minh `A⋮6;31`

   `A=5+5^2+5^3+5^4+…+5^120`

   `⇒A=(5+5^2)+(5^3+5^4)+…+(5^119+5^120)`

   `⇒A=5(1+5)+5^3(1+5)+…+5^119(1+5)`

   `⇒A=6.5+6.5^3+…+6.5^119`

   `⇒A=6(5+5^3+…+5^119)⋮6`

   `A=5+5^2+5^3+5^4+…+5^120`

   `⇒A=(5+5^2+5^3)+(5^4+5^5+5^6)+…+(5^118+5^119+5^120)`

   `⇒A=5(1+5+5^2)+5^4(1+5+5^2)+…+5^118(1+5+5^2)`

   `⇒A=31.5+31.5^4+…+31.5^118`

   `⇒A=6(5+5^4+…+5^118)⋮31`

   `⇒A⋮186` `(đpcm)`

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