198. Calculate the sum:

$$
S=(1+1^2)+(2+2^2)+(3+3^2)+\cdots+(n+n^2)
$$
$\triangle$ Let's break the sum $S$ into two sums, and then apply formulas (1) and (2):

$$
\begin{aligned}
& S=(1+2+3+\cdots+n)+\left(1^{2}+2^{2}+3^{2}+\cdots+n^{2}\right)= \\
& =\frac{n(n+1)}{2}+\frac{n(n+1)(2 n+1)}{6}=\frac{1}{6}(3 n(n+1)+(n(n+1)(2 n+1))= \\
& =\frac{1}{6} n(n+1)(2 n+4)=\frac{1}{3} n(n+1)(n+2)
\end{aligned}
$$

Answer: $S=n(n+1)(n+2) / 3$.