cho m= 3^2+3^3+………….+3^99+3^100 chia hết cho 40

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1. @py

   `\text{Mong bạn cho mình ctlhn ạ!!!}`

   `m= 3^2+3^3+………….+3^99+3^100`

   `⇔m=(3^2+3^3+3^4+3^5)+(3^6+3^7+3^8+3^9)+…..+(3^(97)+3^(98)+3^(99)+3^(100))`

   `⇔m=3^2 .( 1+3+9+27)+3^6 .( 1+3+9+27)+…+ 3^(97) .( 1+3+9+27)`

   `⇔m=3^2×40 +3^6×40+…+3^(97)×40`

   `⇔m=40×(3^2+3^6+…+3^(97))\vdots40ĐPCM`

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2. Đáp án:

   `M\vdots40`

   Giải thích các bước giải:

   `M=3^2+3^3+…+3^(99)+3^(100)`

   `->M=(3^2+3^3+3^4+3^5)+(3^6+3^7+3^8+3^9)+…+(3^(97)+3^(98)+3^(99)+3^(100))`

   `->M=3^2(1+3+3^2+3^3)+3^6(1+3+3^2+3^3)+…+3^(97)(1+3+3^2+3^3)`

   `->M=3^2 . 40+3^6 . 40+…+3^(97).40`

   `->M=(3^2+3^6+…+3^(97)).40\vdots40`

   `->M\vdots40`

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