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M = \(1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{2015^2}-\dfrac{1}{2016^2}\)
M = \(1-\dfrac{1}{2016^2}=1-\dfrac{1}{4064256}=\dfrac{4064255}{4064256}\)
Alone selected this answer. -
FC Alan Walker 25/02/2018 at 14:34
We have: \(\dfrac{2n+1}{\left(n^2+n\right)^2}=\dfrac{\left(n+1\right)^2-n^2}{n^2\left(n+1\right)^2}=\dfrac{1}{n^2}-\dfrac{1}{\left(n+1\right)^2}\)
So: \(M=\dfrac{2.1+1}{\left(1^2+1\right)^2}+\dfrac{2.2+1}{\left(2^2+2\right)^2}+...+\dfrac{2.2015+1}{\left(2015^2+2015\right)^2}\)
\(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{2015^2}-\dfrac{1}{2016^2}\)
\(=1-\dfrac{1}{2016^2}< 1\)
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FA Liên Quân Garena 08/01/2018 at 21:57
M = 1−122+122−132+...+120152−120162
M = 1−120162=1−14064256=40642554064256
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So M < 1