Suggestion: Let $a=\sqrt{2 n+1}$ and $b=$ $\sqrt{2 n-1}$.

Facts that Help: Use the identity

$\left(a^{2}+a b+b^{2}\right)(a-b)=a^{3}-b^{3}$.
For a positive integer $n$ consider the function

$$
f(n)=\frac{4 n+\sqrt{4 n^{2}-1}}{\sqrt{2 n+1}+\sqrt{2 n-1}}
$$

Calculate the value of

$$
f(1)+f(2)+f(3)+\cdots+f(40)
$$
Let $a=\sqrt{2 n+1}$ and $b=\sqrt{2 n-1}$. Then $a b=\sqrt{4 n^{2}-1}$, $a^{2}+b^{2}=4 n$ and $a^{2}-b^{2}=2$. Therefore,

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

Since $a-b \neq 0$, we can write

$f(n)=\frac{a^{2}+b^{2}+a b}{a+b} \cdot \frac{a-b}{a-b}=\frac{a^{3}-b^{3}}{a^{2}-b^{2}}=\frac{(\sqrt{2 n+1})^{3}-(\sqrt{2 n-1})^{3}}{2}$.

Thus,

$$
\begin{gathered}
f(1)+f(2)+\cdots+f(40)= \\
=\frac{(\sqrt{3})^{3}-(\sqrt{1})^{3}}{2}+\frac{(\sqrt{5})^{3}-(\sqrt{3})^{3}}{2}+\cdots+\frac{(\sqrt{81})^{3}-(\sqrt{79})^{3}}{2}= \\
=\frac{(\sqrt{81})^{3}-(\sqrt{1})^{3}}{2}=\frac{729-1}{2}=364
\end{gathered}
$$