chứng tỏ S=1/2.2+1/2.2.2+1/2.2.2.2+…+1/2.mũ 100<1

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1. `S= 1/2^2 + 1/2^3 + 1/2^4 +….+ 1/2^100`

   `1/2 S  = 1/2( 1/2^2 + 1/2^3 + 1/2^4 +…+ 1/2^100)`

   `1/2 S= 1/2^3 + 1/2^4 + 1/2^5 +…+ 1/2^101`

   `S- 1/2S = 1/2^2+ 1/2^3 + 1/2^4 +…+ 1/2^100 – 1/2^3 – 1/2^4- 1/2^5 -…- 1/2^101`

   `1/2 S= 1/2^2 – 1/2^101`

   `S= (1/2^2 – 1/2^101) : 1/2`

   `S= (1/2^2 – 1/2^101) . 2`

   `S= 1/2 – 1/2^100`

   Vì `1/2< 1`

   `=> 1/2 – 1/2^100 < 1`

   Vậy `S < 1`

    

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2. `S = 1/(2.2) + 1/(2.2.2) + 1/(2.2.2.2) + … + 1/2^100`

   `⇒ S = 1/2^2 + 1/2^3 + 1/2^4 + … + 1/2^100`
   `⇒ 2S = 2. (1/2^2 + 1/2^3 + 1/2^4 + … + 1/2^100)`
   

   `⇒ 2S = 1/2 + 1/2^2 + 1/2^3 + … + 1/2^99`

   `⇒ 2S – S = (1/2 + 1/2^2 + 1/2^3 + … + 1/2^99) – (1/2^2 + 1/2^3 + 1/2^4 + … + 1/2^100)`

   `⇒ S = 1/2 – 1/2^100`

   `Vì  1/2 < 1 ⇒ 1/2 – 1/2^100 < 1`

   `⇒ S < 1`

   `⇒ đpcm`

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