Question

Cho: S=\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{2018^2}\)chứng tỏ S< \(\dfrac{1}{4}\)

Answer 1

S=\(\dfrac{1}{5.5}+\dfrac{1}{6.6}+\dfrac{1}{7.7}+...+\dfrac{1}{2018.2018}\)

Ta có: \(\dfrac{1}{5.5}< \dfrac{1}{4.5};\dfrac{1}{6.6}< \dfrac{1}{5.6};\dfrac{1}{7.7}< \dfrac{1}{6.7};...;\dfrac{1}{2018.2018}< \dfrac{1}{2017.2018}\)

\(\Rightarrow\) S<\(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{2017.2018}\)

S<\(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\)

S< \(\dfrac{1}{4}-\dfrac{1}{2018}< \dfrac{1}{4}\)

\(\Rightarrow\)S<\(\dfrac{1}{4}\)

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