S=1/2^2 – 1/2^4 +1/2^6 -..+ 1/2^4n-2 – 1/2^4n +…+ 1/2^2002 – 1/2^2004 CMR S<0,2

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1. Đáp án + giải thích bước giải :

   `S = 1/2^2 – 1/2^4 + 1/2^6 – … + 1/2^{4n – 2} – 1/2^{4n} + … + 1/2^{2002} – 1/2^{2004}`

   `-> 2^2S = 1  -1/2^2 + 1/2^4 + … + 1/2^{4n  -4} – 1/2^{4n  -2} + … + 2/2^{2000} – 1/2^{2002}`

   `-> 4S + S = (1  -1/2^2 + 1/2^4 + … + 1/2^{4n  -4} – 1/2^{4n  -2} + … + 2/2^{2000} – 1/2^{2002}) + (1/2^2 – 1/2^4 + 1/2^6 – … + 1/2^{4n – 2} – 1/2^{4n} + … + 1/2^{2002} – 1/2^{2004})`

   `-> 5S = 1 – 1/2^{2004}`

   Ta thấy : `1 – 1/2^{2004} < 1`

   `-> 5S = 1 – 1/2^{2004} < 1`

   `-> S = 1 – 1/2^{2004} < 1/5`

   `-> S = 1 – 1/2^{2004} < 0,2`

   hay `S < 0,2 (đpcm)`

    

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2. Giải thích các bước giải:

   Ta có:

   $S=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-…+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+…+\dfrac{2}{2^{2002}}-\dfrac{1}{2^{2004}}$

   $\to 2^2S=1-\dfrac{1}{2^2}+\dfrac{1}{2^4}-…+\dfrac{1}{2^{4n-4}}-\dfrac{1}{2^{4n-2}}+…+\dfrac{2}{2^{2000}}-\dfrac{1}{2^{2002}}$

   $\to S+2^2S=1-\dfrac{1}{2^{2004}}$

   $\to 5S=1-\dfrac{1}{2^{2004}}<1$

   $\to S<\dfrac15$

   $\to S<0.2$

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