Find ∑((2k+1)+√k(k+[1)])/(√k+√(k+1))

[Albert1]

$$\sum_{k=1}^{99}\dfrac {(2k+1)+\sqrt{k(k+1)}}{\sqrt k+\sqrt{k+1}}$$

 

[kaliprasad]

$$\sum_{k=1}^{99}\dfrac {(2k+1)+\sqrt{k(k+1)}}{\sqrt k+\sqrt{k+1}}$$

Spoiler

we have $(\sqrt{k+1} + \sqrt{k})^2 = 2k + 1 + 2 \sqrt{k(k+1)}$
hence $2k + 1 + \sqrt{k(k+1)} =(\sqrt{k+1} + \sqrt{k})^2 - \sqrt{k(k+1)}$
or $\frac{2k + 1 + \sqrt{k(k+1)}}{\sqrt{k+1} + \sqrt{k}}=(\sqrt{k+1} + \sqrt{k}) - \sqrt{k(k+1)}(\sqrt{k+1}-\sqrt{ k})$
$= (\sqrt{k+1} + \sqrt{k}) - (k+1) \sqrt{k} + k \sqrt{k+1}$
$= (k+1)\sqrt{k+1} - k\sqrt{k}$

this is telescopic sm and adding from 1 to 99 we get the sum =$100 * \sqrt{100} -1 * \sqrt{1} = 999$