We have, $A=\left[\begin{array}{lll}1& 1& 1\\ 0& 1& 1\\ 0& 0& 1\end{array}\right]$ and $M=A+A^2+A^3+\dots +A^{20}$

Now, $A^2=A.A=\left[\begin{array}{lll}1& 1& 1\\ 0& 1& 1\\ 0& 0& 1\end{array}\right]\left[\begin{array}{lll}1& 1& 1\\ 0& 1& 1\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{lll}1& 2& 3\\ 0& 1& 2\\ 0& 0& 1\end{array}\right]$

$\Rightarrow A^3=A^2.A=\left[\begin{array}{lll}1& 1& 1\\ 0& 1& 1\\ 0& 0& 1\end{array}\right]\left[\begin{array}{lll}1& 2& 3\\ 0& 1& 2\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{lll}1& 3& 6\\ 0& 1& 3\\ 0& 0& 1\end{array}\right]$

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$A^n=\left[\begin{array}{ccc}1& n& \frac{n\left(n+1\right)}{2}\\ 0& 1& n\\ 0& 0& 1\end{array}\right]$

So, required sum

$=20\times 3+2\underset{r=1}{\overset{20}{\sum r}}+\sum _{r=1}^{20}\left(\frac{r\left(r+1\right)}{2}\right)$

$=20\times 3+2\underset{r=1}{\overset{20}{\sum r}}+\frac{1}{2}\sum _{r=1}^{20}{r}^2+\frac{1}{2}\sum _{r=1}^{20}r$

$=20\times 3+\frac{5}{2}\sum _{r=1}^{20}r+\frac{1}{2}\sum _{r=1}^{20}{r}^2$

$=20\times 3+\frac{5}{2}\left(\frac{20\times 21}{2}\right)+\frac{1}{2}\left(\frac{20\times 21\times 41}{6}\right)$

$\left[\because \sum _{r=1}^{n}r=\frac{n\left(n+1\right)}{2}, \sum _{r=1}^n{r}^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right]$

$=60+5\left(5\times 21\right)+\left(5\times 7\times 41\right)$

$=60+525+1435$

$=2020$