Đặt \(S=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...-\dfrac{1}{100^2}=1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}-...+\dfrac{1}{100^2}\right)\)
Ta thấy :
\(\dfrac{1}{2^2}>\dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}>\dfrac{1}{2.3}\)
\(\dfrac{1}{3^2}>\dfrac{1}{3.4}\)
\(...\dfrac{1}{100^2}>\dfrac{1}{99.100}\)
\(\Rightarrow S>1-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{99.100}\right)\)
\(\Rightarrow S>1-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(\Rightarrow S>1-1+\dfrac{1}{100}=\dfrac{1}{100}\left(đpcm\right)\)
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