a(x)=x+x^2+x^3+… +x^99+x^100 tại x= 1/2

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1. Cho `x= 1/2`

   `=> A(1/2) = 1/2 + 1/2^2 + 1/2^3 +…+1/2^99 + 1/2^100`

   `=>1/2 A(1/2)= 1/2 ( 1/2 + 1/2^2 +1/2^3 +…+1/2^100)`

   `=> 1/2 A(1/2) = 1/2^2 +1/2^3+ 1/2^4 +…+1/2^101`

   `=> A(1/2) – 1/2A (1/2) = 1/2 + 1/2^2+ 1/2^3 +…+1/2^100 – 1/2^2 – 1/2^3 – 1/2^3-…-1/2^101`

   `=> 1/2A(1/2)= 1/2 – 1/2^101`

   `=> A(1/2) = (1/2 -1/2^101):1/2`

   `=> A(1/2) = (1/2 -1/2^101) . 2`

   `=> A(1/2) = 1 – 1/2^100`

   Vậy `A(1/2) = 1 – 1/2^100`

    

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2. Đáp án :

   Với `x = 1/2`

   `⇔ A (1/2) = 1/2 + (1/2)^2 + (1/2)^3 + … + (1/2)^{99} + (1/2)^{100}`

   `⇔ A(1/2) = 1/2 + 1/2^2 + 1/2^3 + … + 1/2^{99} + 1/2^{100}`

   `⇔ 1/2 A (1/2) = 1/2 (1/2 + 1/2^2 + 1/2^3+…+1/2^{99} + 1/2^{100})`

   `⇔ 1/2 A (1/2) = 1/2^2 + 1/2^3 + 1/2^4 + … + 1/2^{100} + 1/2^{101} (1)`

   $\\$

   Đặt `B = 1/2^2 + 1/2^3 + 1/2^4 + … + 1/2^{100} + 1/2^{101}`

   `⇔ 2 B = 1/2 + 1/2^2 + 1/2^3 + … + 1/2^{99} + 1/2^{100}`

   `⇔ 2B – B = (1/2 + 1/2^2 + 1/2^3 + … + 1/2^{99} + 1/2^{100}) – (1/2^2 + 1/2^3 + 1/2^4 + … + 1/2^{100} + 1/2^{101})`

   `⇔ B = 1/2 – 1/2^{101} (2)`

   $\\$
   Thay `(2)` vào `(1)` ta được :

   `1/2 A (1/2) = 1/2 – 1/2^{101}`

   `⇔ A(1/2) = (1/2 – 1/2^{101} ) : 1/2`

   `⇔ A (1/2) = 1 – 1/2^{100}`

   Vậy `A (1/2) = 1 – 1/2^{100}`

    

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