Cho B= (1−1/2^2)⋅(1−1/3^2)⋅(1−1/4^2)⋅…⋅(1−1/2021^2). So sánh B với 1/2

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1. Đáp án: $B>\dfrac12$

   Giải thích các bước giải:

   Ta có:

   $B=(1-\dfrac{1}{2^2})(1-\dfrac{1}{3^2})(1-\dfrac1{4^2})…(1-\dfrac{1}{2021^2})$

   $\to B=\dfrac{2^2-1}{2^2}.\dfrac{3^2-1}{3^2}.\dfrac{4^2-1}{4^2}…\dfrac{2021^2-1}{2021^2}$

   $\to B=\dfrac{(2-1)(2+1)}{2^2}.\dfrac{(3-1)(3+1)}{3^2}.\dfrac{(4-1)(4+1)}{4^2}…\dfrac{(2021-1)(2021+1)}{2021^2}$

   $\to B=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}…\dfrac{2020.2022}{2021^2}$

   $\to B=\dfrac{1.2.3…2020}{2.3.4..2021}.\dfrac{3.4.5..2022}{2.3.4…2021}$

   $\to B=\dfrac1{2021}.\dfrac{2022}{2}$

   $\to B=\dfrac{2022}{2021}.\dfrac{1}{2}>1.\dfrac12$

   $\to B>\dfrac12$

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2. Đáp án + giải thích bước giải :

   `B = (1 – 1/2^2) (1 – 1/3^2) (1 – 1/4^2) ….. (1-1/2021^2)`

   `-> B = (1 . 3)/2^2 . (2 . 4)/3^2 . (3.5)/4^2…. (2020 . 2020)/2021^2`

   `-> B = (1 . 3 ….. 2020)/(2 . 3 . 4 ….. 2021) . (3 . 4 . 5. …. 2022)/(2 . 3 . 4 …. 2021)`

   `-> B = 1/2021 . 2020/2`

   Ta thấy : `1/2021 . 2020/2 = 2022/2021 . 1/2 > 1/2`

   `-> B > 1/2` 

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