Radicals are getting crazy in 2016

Let the sum be $S$ , then:

$\begin{aligned} S & = \sum_{n=1}^{2016} \frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}} \\ & = \sum_{n=1}^{2016} \frac{(n+1)\sqrt{n} - n\sqrt{n+1}}{((n+1)\sqrt{n} + n\sqrt{n+1})((n+1)\sqrt{n} - n\sqrt{n+1}} \\ & = \sum_{n=1}^{2016} \frac{(n+1)\sqrt{n} - n\sqrt{n+1}}{n(n+1)^2 - n^2(n+1)} \\ & = \sum_{n=1}^{2016} \frac{(n+1)\sqrt{n} - n\sqrt{n+1}}{n(n+1)(n+1 - n)} \\ & = \sum_{n=1}^{2016} \frac{(n+1)\sqrt{n} - n\sqrt{n+1}}{n(n+1)} \\ & = \sum_{n=1}^{2016} \left( \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} \right) \\ & = \sum_{n=1}^{2016} \frac{1}{\sqrt{n}} - \sum_{n=2}^{2017} \frac{1}{\sqrt{n}} \\ & = \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{2017}} \\ & = \frac{\sqrt{2017} -1} {\sqrt{2017}} \end{aligned}$

$\Rightarrow a = \boxed{2017}$

Yay, rohit followed me!

I'm writing a solution on the first problem on his feed XD First, rationalize the denominator

$\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}} = \frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}} \cdot \frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{(n+1)\sqrt{n}-n\sqrt{n+1}} = \frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{n^2+n} = \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}$

This makes the sum above telescope to $\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+ \cdots +\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}$

$= 1-\frac{1}{\sqrt{2017}} = \frac{\boxed{\sqrt{2017}}-1}{\boxed{\sqrt{2017}}}$

Therefore, $a=\boxed{2017}$

$\ \sum _{ n=1 }^{ 2016 }{ \frac { 1 }{ n\sqrt { n+1 } +\sqrt { n } (n+1) } } =\sum _{ n=1 }^{ 2016 }{ \frac { { (\sqrt { n+1 } ) }^{ 2 }-{ (\sqrt { n } ) }^{ 2 } }{ \sqrt { n } \cdot \sqrt { n+1 } (\sqrt { n } +\sqrt { n+1 } ) } } =\sum { \frac { \sqrt { n+1 } -\sqrt { n } }{ \sqrt { n } .\sqrt { n+1 } } } \\ =\sum { \frac { 1 }{ \sqrt { n } } } -\frac { 1 }{ \sqrt { n+1 } } =\quad \frac { 1 }{ 1 } -\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } } -\frac { 1 }{ \sqrt { 3 } } ..........-\frac { 1 }{ \sqrt { 2017 } } =1-\frac { 1 }{ \sqrt { 2017 } } =\frac { \sqrt { 2017 } -1 }{ \sqrt { 2017 } } \\ \Rightarrow \boxed { a=2017 }$

Using fractional decomposition:

$\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}$

Then it's a telescoping sequence.

Rationalize the given expresses ion and convert into 1 by root n minus 1 by root n+1 form ...to cancel terms....always aim to simplify the exp into a form in which terms cancel out while performing the given operation..(here it is addition..)