Đáp án:
$S < \dfrac37$
Giải thích các bước giải:
Với $\forall n \geqslant 0$ ta có:
$\quad n + (n+1) > 2\sqrt{n(n+1)}\quad (BDT\ Cauchy)$
$\Leftrightarrow 2n + 1 > 2\sqrt{n(n+1)}$
$\Leftrightarrow \dfrac{1}{2n+1} < \dfrac{1}{2\sqrt{n(n+1)}}$
$\Leftrightarrow \dfrac{1}{(2n+1)\left(\sqrt{n} +\sqrt{n+1}\right)} < \dfrac{1}{2\sqrt{n(n+1)}\left(\sqrt{n} +\sqrt{n+1}\right)}$
$\Leftrightarrow \dfrac{1}{(2n+1)\left(\sqrt{n} +\sqrt{n+1}\right)} < \dfrac{\sqrt{n+1} - \sqrt n}{2\sqrt{n(n+1)}\left(\sqrt{n} +\sqrt{n+1}\right)\left(\sqrt{n+1} - \sqrt n\right)}$
$\Leftrightarrow \dfrac{1}{(2n+1)\left(\sqrt{n} +\sqrt{n+1}\right)} < \dfrac{\sqrt{n+1} -\sqrt n}{2\sqrt{n(n+1)}}$
$\Leftrightarrow \dfrac{1}{(2n+1)\left(\sqrt{n} +\sqrt{n+1}\right)} < \dfrac12\left(\dfrac{1}{\sqrt n} -\dfrac{1}{\sqrt{n+1}}\right)$
Áp dụng:
$\quad S = \dfrac{1}{3\left(1+\sqrt2\right)} +\dfrac{1}{5\left(\sqrt2+\sqrt3\right)}+ \cdots + \dfrac{1}{97\left(\sqrt{48}+\sqrt{49}\right)}$
$\Leftrightarrow S< \dfrac12\left(\dfrac{1}{\sqrt1} -\dfrac{1}{\sqrt2}\right) + \dfrac12\left(\dfrac{1}{\sqrt2} -\dfrac{1}{\sqrt3}\right)+\cdots + \dfrac12\left(\dfrac{1}{\sqrt{48}} -\dfrac{1}{\sqrt{49}}\right)$
$\Leftrightarrow S < \dfrac12\left(1 - \dfrac{1}{7}\right)$
$\Leftrightarrow S < \dfrac37$
Vậy $S < \dfrac37$
