[Toán 8] Nâng cao đại số cần giải gấp -_-

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promantickg2000

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nguyenbahiep1

[laTEX]\frac{1}{2^2} < \frac{1}{1.2} = 1 - \frac{1}{2} \\ \\ \frac{1}{3^2} < \frac{1}{2.3} = \frac{1}{2} - \frac{1}{3} \\ \\ \frac{1}{4^2} < \frac{1}{3.4} = \frac{1}{3} - \frac{1}{4} \\ \\ ............................. \\ \\ \frac{1}{n^2} < \frac{1}{(n-1).n} = \frac{1}{n-1} - \frac{1}{n} \\ \\ S = \frac{1}{2^2}+ \frac{1}{3^2} + ...+\frac{1}{n^2} <1 - \frac{1}{2} + \frac{1}{2}- \frac{1}{3} + .. - \frac{1}{n} \\ \\ S < 1 - \frac{1}{n} < 1 \Rightarrow S < 1 \Rightarrow dpcm[/laTEX]
 
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