C=1+3+3^2+…+3^23 a) Rút gọn b) C/m C chia hết cho 13 c) C/m C chia hết cho 40 cứuuuuuuuuuuuuuuuuuuu

2 bình luận về “C=1+3+3^2+…+3^23 a) Rút gọn b) C/m C chia hết cho 13 c) C/m C chia hết cho 40 cứuuuuuuuuuuuuuuuuuuu”

1. a)C=1+3+3^2+…+3^23

   =>3C=3+3^2+3^3+…+3^24

   =>3C-C=(3+3^2+3^3+…+3^24)-(1+3+3^2+…+3^23)

   =>3C-C=3+3^2+3^3+…+3^24-1+3+3^2-…-3^23

    =>2C=3^24-1

   =>C=(3^24-1)/2

   b)C=1+3+3^2+…+3^23

   =(1+3+3^2)+…+(3^21+3^22+3^23)

   =(1+3+3^2)+…+3^21(1+3+3^2)

   =(1+3+3^2)(1+…+3^21)

   =13.(1+…+3^21)vdots13

   =>Cvdots13(đpcm)

   c)C=1+3+3^2+…+3^23

   =(1+3+3^2+3^3)+…+(3^20+3^21+3^22+3^23)

   =(1+3+3^2+3^3)+…+3^20(1+3+3^2+3^3)

   =(1+3+3^2+3^3)(1+…+3^20)

   =40.(1+…+3^20)vdots40

   =>Cvdots40(đpcm)

   Trả lời
2. a, Ta có: C = 1 + 3 + 3^2 + … + 3^(23)

   => 3C = 3 + 3^2  + 3^3 + … + 3^(24)

   => 3C – C = (3 + 3^2 + 3^3 + …+ 3^(24)) – (1 + 3 + 3^2 + … + 3^(23))

   => 2C  = 3^(24) – 1

   => C = (3^(24) – 1)/2

   Vậy C = (3^(24) – 1)/2

   b, Ta có: C = 1 + 3 + 3^2 + … + 3^(23)

   => C = (1 + 3 + 3^2) + …. + (3^(21) + 3^(22) + 3^(23))

   => C = (1 + 3 + 3^2) + … + 3^(21) (1 + 3 + 3^2)

   => C = 13 + …. + 3^(21) . 13

   => C = 13(1 + … + 3^(21)) \vdots 13

   => C \vdots 13  (đpcm)

   c, Ta có: C = 1 + 3 + 3^2 + … + 3^(23)

   => C = (1 + 3 + 3^2 + 3^3) +( 3^4 +3^5 + 3^6 + 3^7) + … + ((3^(16) + 3^(17) + 3^(18) + 3^(19)) + (3^(20) + 3^(21) + 3^(22) + 3^(23)))

   => C = (1 + 3 + 3^2 + 3^3) + 3^4 (1 + 3 + 3^2 + 3^3) + … + 3^(16) (1 + 3 + 3^2 + 3^3) + 3^(20) (1 + 3 + 3^2 + 3^3)

   => C = 40 + 3^4 . 40… + 3^(16) . 40 + 3^(20) . 40

   => C = 40 (1 + 3^4 +.. + 3^(16) + 3^(20)) \vdots 40

   => C \vdots 40  (đpcm)

   $#duong612009$

   Trả lời