[Toán 7] Nâng cao

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iceghost

Bài 2

$B=(1-\dfrac{1}{2^2})(1-\dfrac{1}{3^2})(1-\dfrac{1}{4^2})...(1-\dfrac{1}{n^2})$

$=(1-\dfrac{1}{2})(1+\dfrac{1}{2})(1-\dfrac{1}{3})(1+\dfrac{1}{3})(1-\dfrac{1}{4})(1+\dfrac{1}{4})...(1-\dfrac{1}{n})(1+\dfrac{1}{n})$

$=\dfrac{1}{2}.\dfrac{3}{2}.\dfrac{2}{3}.\dfrac{4}{3}.\dfrac{3}{4}.\dfrac{5}{4}. ... .\dfrac{n-1}{n}.\dfrac{n+1}{n}$

$=\dfrac12 . \dfrac{n+1}{n}=\dfrac{n+1}{2n}$
 
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