Question

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

Answer 1

Answer: 364

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