b1: tính nhanh A=1+1/3+1/3^2+…..+1/3^n B=1/2+1/2^2+1/2^3+…+1/2^100

2 bình luận về “b1: tính nhanh A=1+1/3+1/3^2+…..+1/3^n B=1/2+1/2^2+1/2^3+…+1/2^100”

1. Ta có: A = 1 + 1/3 + 1/(3^2) + … + 1/(3^n)

   => 3A = 3 + 1 + 1/3 + … + 1/(3^(n- 1))

   => 3A – A = ( 3 + 1 + 1/3 + … + 1/(3^(n- 1))) – (1 + 1/3 + 1/(3^2) + … + 1/(3^n))

   => 2A = 3 – 1/(3^n)

   => A = (3 – 1/(3^n))/2

   Vậy A = (3 – 1/(3^n))/2

   ————————————-

   Ta có: B = 1/2 + 1/(2^2) + 1/(2^3) + … + 1/(2^(100))

   => 2B = 1 + 1/2 + 1/(2^2) + … + 1/(2^(99))

   => 2B – B = (1 + 1/2 + 1/(2^2) + … + 1/(2^(99))) – (1/2 + 1/(2^2) + 1/(2^3) + … + 1/(2^(100)))

   => B = 1 – 1/(2^(100))

   Vậy B = 1 – 1/(2^(100))

   $#duong612009$

   Trả lời
2. Gửi e 

   B = 1/2 + 1/(2^2) + 1/(2^3) + … + 1/(2^(100))

   => 2B = 1 + 1/2 + 1/(2^2) + … + 1/(2^(99))

   => 2B – B = (1 + 1/2 + 1/(2^2) + … + 1/(2^(99))) – (1/2 + 1/(2^2) + 1/(2^3) + … + 1/(2^(100)))

   => B = 1 – 1/(2^(100))

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