Weird function

$\begin{aligned}f(n)&=\dfrac{4n+\sqrt{4n^2-1}}{\sqrt{2n+1}+\sqrt{2n-1}}\\&=\dfrac{2n+1+2n-1+\sqrt{2n+1}\sqrt{2n-1}}{\sqrt{2n+1}+\sqrt{2n-1}} \\&=\dfrac{\left( 2n+1+2n-1+\sqrt{2n+1}\sqrt{2n-1}\right)\left( \sqrt{2n+1}-\sqrt{2n-1} \right) }{\left(\sqrt{2n+1}+\sqrt{2n-1}\right)\left( \sqrt{2n+1}-{\sqrt{2n-1}} \right)}\\&=\dfrac{{(2n+1)}^{\frac{3}{2}}-{(2n-1)}^{\frac{3}{2}}}{2}\end{aligned}$
$\begin{aligned}\displaystyle\sum_{k=1}^{40} f(k)&=\frac{1}{2}\displaystyle\sum_{k=1}^{40} {(2k+1)}^{\frac{3}{2}}-{(2k-1)}^{\frac{3}{2}}\\&=\frac{1}{2}\left( {81}^{\frac{3}{2}}-{1}^{\frac{3}{2}}\right)\\&=\frac{1}{2}\left( 729-1\right)\\&\huge\color{#3D99F6}{=\boxed{364}}\end{aligned}$

$Let~a=\sqrt{2n+1},~~~~~b=\sqrt{2n-1}.\\ So~~~a+b=\sqrt{2n+1}+\sqrt{2n-1},~~~~~a-b=\sqrt{2n+1}-\sqrt{2n-1}.~~~~~ab=\sqrt{4n^2-1},\\ a^2=2n+1,~~~~~b^2=2n-1,~~~~~~~~a^2+b^2=4n,~~~~~(a+b)*(a-b)=2.\\ \implies~f(n)=\dfrac {a^2+b^2+ab } {a+b}=\dfrac {~~(a^2+b^2+ab)~*~(a-b) } {(a+b)*(a-b)}.\\ \displaystyle \sum_{n=1}^{40}\dfrac{a^3-b^3} 2=\frac 1 2 \sum_{n=1}^{40}(2n+1)*\sqrt{2n+1}-\frac 1 2 \sum_{n=1}^{40}(2n-1)*\sqrt{2n-1}\\$ $\displaystyle=\frac 1 2 \sum_{n=1}^{39}(2n+1)*\sqrt{2n+1}+\frac 1 2 \sum_{n=40}^{40}(2n+1)*\sqrt{2n+1}-\frac 1 2 \sum_{n=1}^{1}(2n-1)*\sqrt{2n-1}-\frac 1 2 \sum_{n=2}^{40}(2n-1)*\sqrt{2n-1}\\ \displaystyle=\frac 1 2 \sum_{n=1}^{39}(2n+1)*\sqrt{2n+1}+\frac 1 2 (81\sqrt{81}-1\sqrt1)-\frac 1 2 \sum_{n=1}^{39}(2n+1)*\sqrt{2n+1}\\ \displaystyle \sum_{n=1}^{40}f(n)=\frac 1 2*(729 - 1)=\Large~~\color{#D61F06}{364}.$

Mutiply by sqrt(2n+1)-sqrt(2n-1) in numerator and denominator => simplifying denominator leads to 2, simplyfying numerator leads to (2n+1)^3/2 - (2n-1)^3/2..

For summation 1 to 40 => (81^3/2-1^3/2) / 2 = 364