Question

Bài 12: Cho B = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + ... + \(\dfrac{1}{9^2}\)

CMR: B > \(\dfrac{2}{5}\)

Giúp mk vs, ai nhanh mk sẽ tick

Answer 1

Tacó :

B = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{9^2}\) \(\Rightarrow\)Đặt D=\(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\)<B

\(\Rightarrow\)D= \(\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{3}-.....+\dfrac{1}{9}-\dfrac{1}{10}\) \(\Rightarrow D=\dfrac{1}{2}-\dfrac{1}{10}\)

\(\Rightarrow D=\dfrac{2}{5}\)

Vì D =\(\dfrac{2}{5}\) =\(\dfrac{2}{5}\)

mà D<B

\(\Rightarrow\)B>\(\dfrac{2}{5}\)(dpcm)

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