If A=[ … ] and M=A+A^2+A^3+… +A^20, then the sum of all the elements of the matrix M is equal to .

We have, $A=[\begin{matrix}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{matrix}]$A=[100​110​111​] and $M=A+{A}^{2}+{A}^{3}+\ldots +{A}^{20}$M=A+A2+A3+…+A20

Now, ${A}^{2}=A.A=[\begin{matrix}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{matrix}][\begin{matrix}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{matrix}]=[\begin{matrix}1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{matrix}]$A2=A.A=[100​110​111​][100​110​111​]=[100​210​321​]

$$\Rightarrow {A}^{3}={A}^{2}.A=[\begin{matrix}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{matrix}][\begin{matrix}1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{matrix}]=[\begin{matrix}1 & 3 & 6 \\ 0 & 1 & 3 \\ 0 & 0 & 1\end{matrix}]$$⇒A3=A2.A=[100​110​111​][100​210​321​]=[100​310​631​]

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$${A}^{n}=[\begin{matrix}1 & n & \frac{n(n+1)}{2} \\ 0 & 1 & n \\ 0 & 0 & 1\end{matrix}]$$An=[100​n10​2n(n+1)​n1​]

So, required sum

$$=20\times 3+2\sum r_{r=1}^{20}+\sum _{r=1}^{20}(\frac{r(r+1)}{2})$$=20×3+2∑rr=120​+r=1∑20​(2r(r+1)​)

$$=20\times 3+2\sum r_{r=1}^{20}+\frac{1}{2}\sum _{r=1}^{20}{r}^{2}+\frac{1}{2}\sum _{r=1}^{20}r$$=20×3+2∑rr=120​+21​r=1∑20​r2+21​r=1∑20​r

$$=20\times 3+\frac{5}{2}\sum _{r=1}^{20}r+\frac{1}{2}\sum _{r=1}^{20}{r}^{2}$$=20×3+25​r=1∑20​r+21​r=1∑20​r2

$$=20\times 3+\frac{5}{2}(\frac{20\times 21}{2})+\frac{1}{2}(\frac{20\times 21\times 41}{6})$$=20×3+25​(220×21​)+21​(620×21×41​)

$$[\because \sum _{r=1}^{n}r=\frac{n(n+1)}{2},\sum _{r=1}^{n}{r}^{2}=\frac{n(n+1)(2n+1)}{6}]$$[∵r=1∑n​r=2n(n+1)​,r=1∑n​r2=6n(n+1)(2n+1)​]

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

$$=60+525+1435$$=60+525+1435

$$=2020$$=2020