Summation of functions

Substituting $\sqrt { 2n+1 }=a$ and $\sqrt { 2n-1 }=b$ , we have the following property:

$\sqrt { 2n+1 }\cdot \sqrt { 2n-1 } =\sqrt { 4{ n }^{ 2 }-1 }=ab$

$(\sqrt { 2n+1 })^2+(\sqrt { 2n-1 })^2=4n=a^2+b^2$ . Our function becomes:

$f\left( n \right) =\frac{a^2+b^2+ab}{a+b}$

Multiplying by $\frac{a-b}{a-b}$ :

$f\left( n \right) =\frac{a^2+b^2+ab}{a+b}\cdot \frac{a-b}{a-b} = \frac{a^3-b^3}{a^2-b^2}$

Substituting back the values of $a$ and $b$ :

$f\left( n \right) =\frac{(\sqrt { 2n+1 })^3-(\sqrt { 2n-1 })^3}{\sqrt { 2n+1 })^2-(\sqrt { 2n-1 })^2}=\frac{(\sqrt { 2n+1 })^3-(\sqrt { 2n-1 })^3}{2n+1-2n+1 }=\frac{(\sqrt { 2n+1 })^3-(\sqrt { 2n-1 })^3}{2 }$

The sum $f\left( 1 \right)+f\left( 2 \right)+f\left( 3 \right)+...+f\left( 38 \right)+f\left( 39 \right)+f\left( 40 \right)$ becomes:

$\displaystyle{\frac{(\sqrt { 3 })^3-(\sqrt { 1 })^3}{2 }+\frac{(\sqrt { 5 })^3-(\sqrt { 3 })^3}{2 }+\frac{(\sqrt { 7 })^3-(\sqrt { 5 })^3}{2 }+...+\frac{(\sqrt { 77 })^3-(\sqrt { 75 })^3}{2 }+\frac{(\sqrt { 79 })^3-(\sqrt { 77 })^3}{2 }+\frac{(\sqrt { 81 })^3-(\sqrt { 79 })^3}{2 }}$

Canceling equivalent terms we will finish with $\frac{-(\sqrt { 1 })^3+(\sqrt { 81 })^3}{2 }=\frac{-(1)^3+(9)^3}{2 }=\frac{-1+729}{2 }=\boxed{364}$

$\begin{aligned} S & = f(1) + f(2) + f(3) + \cdots + f(40) \\ & = \sum_{n=1}^{40} f(x) \\ & = \sum_{n=1}^{40} \frac {4n+\sqrt{4n^2-1}}{\sqrt{2n-1}+\sqrt{2n+1}} \\ & = \sum_{n=1}^{40} \frac {(2n+1)+2\sqrt{(2n+1)(2n-1)} +(2n-1) -\sqrt{(2n+1)(2n-1)}}{\sqrt{2n+1}+\sqrt{2n-1}} \\ & = \sum_{n=1}^{40} \frac {(\sqrt{2n+1}+\sqrt{2n-1})^2 -\sqrt{(2n+1)(2n-1)}}{\sqrt{2n+1}+\sqrt{2n-1}} \\ & = \sum_{n=1}^{40} \left( \sqrt{2n+1}+\sqrt{2n-1} - \frac {\sqrt{(2n+1)(2n-1)}(\sqrt{2n+1}-\sqrt{2n-1})}{(\sqrt{2n+1}+\sqrt{2n-1})(\sqrt{2n+1}-\sqrt{2n-1})} \right) \\ & = \sum_{n=1}^{40} \left( \sqrt{2n+1}+\sqrt{2n-1} - \frac {(2n+1)\sqrt{2n-1}-(2n-1)\sqrt{2n+1})}{2n+1-2n+1} \right) \\ & = \sum_{n=1}^{40} \left( \sqrt{2n+1}+\sqrt{2n-1} - \frac {(2n+1)\sqrt{2n-1}-(2n-1)\sqrt{2n+1})}2 \right) \\ & = \sum_{n=1}^{40} \left( \sqrt{2n+1}+\sqrt{2n-1} - \left(\frac {2n+1}2 \right) \sqrt{2n-1} + \left(\frac {2n-1}2 \right) \sqrt{2n+1} \right) \\ & = \sum_{n=1}^{40} \left( \frac {\left(\sqrt{2n+1}\right)^3}2 - \frac {\left(\sqrt{2n-1}\right)^3}2 \right) \\ & = - \frac {\left(\sqrt{2(1)-1}\right)^3}2 + \frac {\left(\sqrt{2(40)-1}\right)^3}2 = - \frac 12 + \frac {729}2 = \boxed{364} \end{aligned}$