If A = [ cases 1 & 1 & 1 0 & 1 & 1 0 & 0 & 1 cases ] and M = A + A...

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

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

⇒ {A}^{3} = {A}^{2} . A = \left[\begin{cases}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{cases}\right]\left[\begin{cases}1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{cases}\right] = \left[\begin{cases}1 & 3 & 6 \\ 0 & 1 & 3 \\ 0 & 0 & 1\end{cases}\right]

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{A}^{n} = \left[\begin{cases}1 & n & \dfrac{n\left(n + 1\right)}{2} \\ 0 & 1 & n \\ 0 & 0 & 1\end{cases}\right]

So, required sum

$= 20 \times 3 + 2\mathop{∑ r}_{r = 1}^{20} + \mathop{∑}_{r = 1}^{20}\left(\dfrac{r\left(r + 1\right)}{2}\right)$

$= 20 \times 3 + 2\mathop{∑ r}_{r = 1}^{20} + \dfrac{1}{2}\mathop{∑}_{r = 1}^{20}{r}^{2} + \dfrac{1}{2}\mathop{∑}_{r = 1}^{20}r$

$= 20 \times 3 + \dfrac{5}{2}\mathop{∑}_{r = 1}^{20}r + \dfrac{1}{2}\mathop{∑}_{r = 1}^{20}{r}^{2}$

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

\left[∵ \mathop{∑}_{r = 1}^{n}r = \dfrac{n\left(n + 1\right)}{2}, \mathop{∑}_{r = 1}^{n}{r}^{2} = \dfrac{n\left(n + 1\right)\left(2n + 1\right)}{6}\right]$</p><p>$= 60 + 5\left(5 \times 21\right) + \left(5 \times 7 \times 41\right)$</p><p>$= 60 + 525 + 1435$</p><p>$= 2020