If A=pmatrix1&1&10&1&10&0&1pmatrix and M=A+A^2+A^3++A^20, then the sum of all the elements…

Question

If $A=\begin{pmatrix}1&1&1\0&1&1\0&0&1\end{pmatrix}$ and $M=A+A^2+A^3+\cdots+A^{20}$, then the sum of all the elements of the matrix $M$ is equal to

Answer: 2020

Accepted Answer

$A^n=\begin{pmatrix}1&n&\binom{n+1}2\0&1&n\0&0&1\end{pmatrix}$ (standard pattern for this unipotent upper-triangular matrix).

$$M=\begin{pmatrix}20 & \sum_{n=1}^{20}n & \sum_{n=1}^{20}\binom{n+1}2\0&20&\sum n\0&0&20\end{pmatrix}.$$

$\sum_{n=1}^{20}n=210$. $\sum_{n=1}^{20}\binom{n+1}2=\sum_{n=1}^{20}\dfrac{n(n+1)}2=\dfrac12\left(\dfrac{20\cdot21\cdot41}6+210\right)=\dfrac12(2870+210)=1540$.

Sum of all elements of $M$: $20+210+1540+20+210+20=2020$.

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