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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 6: Cho S=1+3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9. Chứng tỏ rằng S chia hết cho 4

Toán Lớp 6: Cho S=1+3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9. Chứng tỏ rằng S chia hết cho 4

Comments ( 2 )

  1. Lời giải và giải thích chi tiết:
    $S=1+3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9$
    $S=(1+3)+(3^2+3^3)+(3^4+3^5)+(3^6+3^7)+(3^8+3^9)$
    $S=(1+3)+3^2.(1+3)+3^4.(1+3)+3^6.(1+3)+3^8.(1+3)$
    $S=(1+3)(1+3^2+3^4+3^6+3^8)$
    $S=4.(1+3^2+3^4+3^6+3^8)\vdots 4$
    Vậy $S\vdots 4$

  2. S=1+3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9.
    S=(1+3)+(3^2+3^3)+…..+(3^8+3^9)
    S=1.(1+3)+3^2.(1+3)+…+3^8.(1+3)
    S=(1+3^3+…+3^8).(1+3)
    S=(1+3^3+…+3^8).4 $\vdots$ 4
    Vậy S $\vdots$ 4 

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222-9+11+12:2*14+14 = ? ( )