B1:Tính nhanh C=1/2-1/2^2+1/2^3-1/2^4+…+1/2^99-1/2^100

2 bình luận về “B1:Tính nhanh C=1/2-1/2^2+1/2^3-1/2^4+…+1/2^99-1/2^100”

1. Ta có: C = 1/2 – 1/(2^2) + 1/(2^3) – 1/(2^4) + … + 1/(2^(99)) – 1/(2^(100))

   => 2C = 1 – 1/2 + 1/(2^2) – 1/(2^3) + … + 1/(2^(98)) – 1/(2^(99))

   => 2C + C = (1 – 1/2 + 1/(2^2) – 1/(2^3) + … + 1/(2^(98)) – 1/(2^(99))) + (1/2 – 1/(2^2) + 1/(2^3) – 1/(2^4) + … + 1/(2^(99)) – 1/(2^(100)))

   => 3C = 1 – 1/(2^(100))

   => C = (1 – 1/(2^(100)))/3

   Vậy C = (1 – 1/(2^(100)))/3

   $#duong612009$

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2. C=1/2-1/2^2+1/2^3-1/2^4+…+1/2^99-1/2^100

   <=> 2C = 1 – 1/2+1/2^2-1/2^3+1/2^4-…-1/2^99

   <=> 2C + C = (1 – 1/2+1/2^2-1/2^3+1/2^4-…-1/2^99)+ (1/2-1/2^2+1/2^3-1/2^4+…+1/2^99-1/2^100)

   <=> 3C = (1 – 1/2+1/2^2-1/2^3+1/2^4-…-1/2^99 + 1/2-1/2^2+1/2^3-1/2^4+…+1/2^99-1/2^100)

   <=> 3C = 1 – 1/2^100

   <=> C = (1 – 1/2^100)/3

   $\text{@ngquynhchi331}$

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