Question

For any positive $n,$ let $f\left(n\right)=\frac{4 n + \sqrt{4 n^{2} - 1}}{\sqrt{2 n + 1} + \sqrt{2 n - 1}}.$ Then, $ \sum _{n = 1}^{40}f\left(n\right)$ is equal to

Answer 1

Let, $x=\sqrt{2 n + 1}$ and $y=\sqrt{2 n - 1}\Rightarrow x^{2}+y^{2}=4n$
$x^{2}-y^{2}=2$
Also, $xy=\sqrt{4 n^{2} - 1}$
So,
$f\left(n\right)=\frac{x^{2} + y^{2} + x y}{x + y}=\frac{x^{3} - y^{3}}{x^{2} - y^{2}}=\frac{x^{3} - y^{3}}{2}$
$=\frac{1}{2}\left(\left(2 n + 1\right)^{\frac{3}{2}} - \left(2 n - 1\right)^{\frac{3}{2}}\right)$
Substituting $n=1$ to $40,$ we get
$f\left(1\right)=\frac{1}{2}\left(3^{\frac{3}{2}} - 1^{\frac{3}{2}}\right)$
$f\left(2\right)=\frac{1}{2}\left(5^{\frac{3}{2}} - 3^{\frac{3}{2}}\right)$
$f\left(3\right)=\frac{1}{2}\left(7^{\frac{3}{2}} - 5^{\frac{3}{2}}\right)$
$f\left(40\right)=\frac{1}{2}\left(\left(81\right)^{\frac{3}{2}} - \left(79\right)^{\frac{3}{2}}\right)$
$\sum f \left(\right. n \left.\right)_{n = 1}^{40}=\frac{1}{2}\left(\left(81\right)^{\frac{3}{2}} - 1^{\frac{3}{2}}\right)=\frac{729 - 1}{2}=364$
$\Rightarrow \sum _{n = 1}^{40}f\left(n\right)=364$