Đáp án:
$S = 2007 - \dfrac{1}{2007}=\dfrac{4028048}{2007}$
Giải thích các bước giải:
Xét $1 +\dfrac{1}{n^2} + \dfrac{1}{(n+1)^2}$
$= \left(1+\dfrac{1}{n}\right)^2 - \dfrac2n +\dfrac{1}{(n+1)^2}$
$= \left(1+\dfrac{1}{n}\right)^2 -2 \cdot \dfrac{n+1}{n}\cdot \dfrac{1}{n+1} + +\dfrac{1}{(n+1)^2}$
$= \left(1+\dfrac{1}{n}\right)^2 - 2\cdot \left(1+\dfrac1n\right)\cdot \dfrac{1}{n+1} + +\dfrac{1}{(n+1)^2}$
$= \left(1+\dfrac1n -\dfrac{1}{n+1}\right)^2$
Do đó:
$\sqrt{1 +\dfrac{1}{n^2} + \dfrac{1}{(n+1)^2}}$
$=\sqrt{\left(1+\dfrac1n -\dfrac{1}{n+1}\right)^2}$
$=\left|1+\dfrac1n -\dfrac{1}{n+1}\right|$
Áp dụng:
$S =\sqrt{1+\dfrac{1}{1^2} +\dfrac{1}{2^2}} +\sqrt{1+\dfrac{1}{2^2} +\dfrac{1}{3^2}}+\cdots + \sqrt{1+\dfrac{1}{2006^2} +\dfrac{1}{2007^2}}$
$\to S = \left|1+\dfrac{1}{1} -\dfrac{1}{2}\right|+\left|1+\dfrac{1}{2} -\dfrac{1}{3}\right|+\cdots +\left|1+\dfrac{1}{2006} -\dfrac{1}{2007}\right|$
$\to S = 1+\dfrac{1}{1} -\dfrac{1}{2}+1+\dfrac{1}{2} -\dfrac{1}{3}+\cdots +1+\dfrac{1}{2006} -\dfrac{1}{2007}$
$\to S = 2007 - \dfrac{1}{2007}$
$\to S =\dfrac{4028048}{2007}$
