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

Toán Lớp 6: giúp: CM:1+3+3^2+3^3+…+3^101 chia hết cho 4

Toán Lớp 6: giúp:
CM:1+3+3^2+3^3+…+3^101 chia hết cho 4

Comments ( 2 )

  1. Ta có:
    1 + 3 + 3^2 + 3^3 +…+ 3^101
    = (1 + 3) + (3^2 + 3^3) +…+ (3^100 + 3^101)
    = 1(1 + 3) + 3^2(1 + 3) +…+ 3^100(1 + 3)
    = 1.4 + 3^2. 4 + …+ 3^100. 4
    = 4(1 + 3^2 +…+ 3^100) vdots 4
    ⇒ đpcm
     

  2. 1+3+3^2+3^3+…+3^101
    =(1+3)+(3^2+3^3)+…+(3^100+3^101)
    =1(1+3)+3^2(1+3)+…+3^100(1+3)
    =1.4+3^2 .4+…+3^100 .4
    =(1+3^2+…+3^100).4 \vdots 4
    =>1+3+3^2+3^3+…+3^101 \vdots 4

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