Let
$$
S=\frac{1^{2}}{1 \cdot 2}+\frac{1^{2}+2^{2}}{2 \cdot 3}+\frac{1^{2}+2^{2}+3^{2}}{3 \cdot 4}+\ldots .
$$
upto 20 terms
Let $t_{n}$ be $n$th terms of series.
Then, $t_{n}=\frac{1^{2}+2^{2}+3^{2}+\ldots+n^{2}}{n \cdot(n+1)}$
$$
=\frac{\Sigma n^{2}}{n(n+1)}=\frac{n(n+1)(2 n+1)}{6 \cdot n(n+1)}=\frac{2 n+1}{6}
$$
Taking summation on both sides
$$
\begin{aligned}
\sum_{n=1}^{20} t_{n} &=\frac{2}{6} \sum_{n=1}^{20} n+\frac{1}{6} \sum_{n=1}^{20} 1 \\
&=\frac{1}{3} \times\left(\frac{20(20+1)}{2}\right)+\frac{1}{6} \times 20 \\
&=\frac{1}{3}\left(\frac{20 \times 21}{2}\right)+\left(\frac{10}{3}\right)=\frac{210}{3}+\frac{10}{3}=\frac{220}{3}
\end{aligned}
$$