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

Toán Lớp 6: Cho S=3+3^2+3^3+…+3^9 Chứng tỏ rằng S chia hết cho 13

Toán Lớp 6: Cho S=3+3^2+3^3+…+3^9
Chứng tỏ rằng S chia hết cho 13

Comments ( 2 )

  1. $S = 3 + 3^{2} + 3^{3} + … + 3^{9}$
    $S = (3 + 3^{2} + 3^{3}) + (3^{4} + 3^{5} + 3^{6}) + (3^{7} + 3^{8} + 3^{9})$
    $S = 39 + 3^{3} . (3 + 3^{2} + 3^{3}) + 3^{6} . (3 + 3^{2} + 3^{3})$
    $S = 39 + 3^{3} . 39 + 3^{6} . 39$
    $S = 39 . (1 + 3^{3} + 3^{6})$
    \text{→ S chia hết cho 39.}
    \text{→ S chia hết cho 13.}

  2. S = 3 + 3² + 3³ + … + 3^{9}
       = (3+3²+3³) + … + (3^{7} + 3^{8} + 3^{9})
       = 3 . (1 + 3 + 3²)+…+3^{7} . (1+3+3²)
       = 3 . 13 + … + 3^{7} . 13
       = 13 . (3+..+3^{7}) \vdots 13 (đpcm)
     

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