33. Let $f: \mathbb{N} \rightarrow \mathbb{Q}$ be a function, where $\mathbb{N}$ denotes the set of natural numbers, and $\mathbb{Q}$ denotes the set of rational numbers. Suppose that $f(1)=\frac{3}{2}$, and
$$
f(x+y)=\left(1+\frac{y}{x+1}\right) f(x)+\left(1+\frac{x}{y+1}\right) f(y)+x^{2} y+x y+x y^{2}
$$
for all natural numbers $x, y$. Find the value of $f(20)$.
33. Ans: 4305
Letting $y=1$, one gets
$$
f(x+1)=\left(1+\frac{1}{x+1}\right) f(x)+\left(1+\frac{x}{2}\right) \frac{3}{2}+x^{2}+2 x .
$$

Upon rearranging, one gets
$$
\frac{f(x+1)}{x+2}-\frac{f(x)}{x+1}=x+\frac{3}{4} .
$$

Then we have
$$
\begin{aligned}
-\frac{f(n)}{n+1}-\frac{f(n-1)}{n} & =n-1+\frac{3}{4}, \\
\frac{f(n-1)}{n}-\frac{f(n-2)}{n-1} & =n-2+\frac{3}{4}, \\
\vdots & \\
\frac{f(2)}{3}-\frac{f(1)}{2} & =1+\frac{3}{4} .
\end{aligned}
$$

Adding these equalities together, we get
$$
\begin{aligned}
\frac{f(n)}{n+1}-\frac{f(1)}{2} & =1+2+\cdots+(n-1)+\frac{3}{4}(n-1) \\
& =\frac{(n-1) n}{2}+\frac{3}{4}(n-1) .
\end{aligned}
$$

Thus,
$$
f(n)=(n+1)\left[\frac{(n-1) n}{2}+\frac{3}{4}(n-1)+\frac{1}{2} \cdot \frac{3}{2}\right]=\frac{n(n+1)(2 n+1)}{4} .
$$

Hence,
$$
f(20)=\frac{(20)(21)(41)}{4}=4305 .
$$