Đáp án:
$D = \dfrac{100 - \sqrt5}{100}$
Giải thích các bước giải:
Xét $\dfrac{1}{(n+1)\sqrt n + n\sqrt{n+1}}\quad (n\in\Bbb N^*)$
$=\dfrac{1}{\sqrt n.\sqrt{n+1}\left(\sqrt{n+1} + \sqrt n\right)}$
$=\dfrac{\sqrt{n+1} -\sqrt n}{\sqrt n.\sqrt{n+1}\left(\sqrt{n+1} + \sqrt n\right)\left(\sqrt{n+1} -\sqrt n\right)}$
$=\dfrac{\sqrt{n+1} - \sqrt n}{\sqrt n.\sqrt{n+1}}$
$=\dfrac{1}{\sqrt n} -\dfrac{1}{\sqrt{n+1}}$
Áp dụng:
$\quad D = \dfrac{1}{2\sqrt1 +1\sqrt2} +\dfrac{1}{3\sqrt2 + 2\sqrt3} +\cdots +\dfrac{1}{2000\sqrt{1999} + 1999\sqrt{2000}}$
$\to D =\dfrac{1}{\sqrt1} -\dfrac{1}{\sqrt2} + \dfrac{1}{\sqrt2} -\dfrac{1}{\sqrt3} +\cdots +\dfrac{1}{\sqrt{1999}} -\dfrac{1}{\sqrt{2000}}$
$\to D = 1 -\dfrac{1}{\sqrt{2000}}$
$\to D = \dfrac{100 - \sqrt5}{100}$
