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

We have the given matrix

$$A=\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}.$$

First we separate the diagonal (identity) part from the strictly upper-triangular part. Writing $$I$$ for the identity matrix, we observe

$$A=I+U,\qquad\text{where}\qquad U=A-I=\begin{bmatrix}0&1&1\\0&0&1\\0&0&0\end{bmatrix}.$$

Because every entry of $$U$$ that lies on or below the main diagonal is zero, $$U$$ is nilpotent. Let us check the powers of $$U$$ explicitly:

$$U^2 =\begin{bmatrix}0&1&1\\0&0&1\\0&0&0\end{bmatrix} \begin{bmatrix}0&1&1\\0&0&1\\0&0&0\end{bmatrix} =\begin{bmatrix}0\cdot0+1\cdot0+1\cdot0 & 0\cdot1+1\cdot0+1\cdot0 & 0\cdot1+1\cdot1+1\cdot0\\ 0&0&0\\ 0&0&0 \end{bmatrix} =\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix},$$

and

$$U^3 =U^2U =\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix} \begin{bmatrix}0&1&1\\0&0&1\\0&0&0\end{bmatrix} =\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=0.$$

Thus $$U^3=0,$$ so $$U$$ has nilpotency index $$3$$.

For any positive integer $$n$$ we need $$A^n=(I+U)^n.$$ Because $$I$$ commutes with $$U,$$ the usual binomial theorem for commuting matrices applies:

$$ (I+U)^n =\sum_{k=0}^{n}\binom{n}{k}I^{\,n-k}U^{\,k} =I+\binom{n}{1}U+\binom{n}{2}U^{2}+\binom{n}{3}U^{3}. $$

Since $$U^{3}=0,$$ all terms with $$k\ge 3$$ vanish. Therefore

$$A^n=I+nU+\frac{n(n-1)}{2}\,U^{2}.$$

Now we form the required matrix

$$M=A+A^{2}+A^{3}+\dots+A^{20}=\sum_{n=1}^{20}A^{n}.$$

Substituting the above expression for $$A^n$$ we obtain

$$ M=\sum_{n=1}^{20}\Bigl(I+nU+\frac{n(n-1)}{2}U^{2}\Bigr) =\Bigl(\sum_{n=1}^{20}I\Bigr) +\Bigl(\sum_{n=1}^{20}n\Bigr)U +\Bigl(\sum_{n=1}^{20}\frac{n(n-1)}{2}\Bigr)U^{2}. $$

Let us evaluate each scalar sum one by one.

The first sum is simply

$$\sum_{n=1}^{20}I = 20I.$$

For the second sum we use the formula for the sum of the first $$N$$ natural numbers, $$1+2+\dots+N=\dfrac{N(N+1)}{2},$$ with $$N=20$$:

$$\sum_{n=1}^{20}n=\frac{20\cdot21}{2}=210.$$

For the third sum we need

$$\sum_{n=1}^{20}\frac{n(n-1)}{2} =\frac12\sum_{n=1}^{20}(n^{2}-n).$$

We compute the two parts separately.

The formula for the sum of squares is $$1^{2}+2^{2}+\dots+N^{2}=\dfrac{N(N+1)(2N+1)}{6}.$$ Putting $$N=20$$ we get

$$\sum_{n=1}^{20}n^{2}=\frac{20\cdot21\cdot41}{6}=2870.$$

We already know $$\sum_{n=1}^{20}n=210.$$ Therefore

$$ \sum_{n=1}^{20}(n^{2}-n)=2870-210=2660, $$

and hence

$$ \sum_{n=1}^{20}\frac{n(n-1)}{2}=\frac{2660}{2}=1330. $$

Putting these scalar sums back, we have

$$M=20I+210\,U+1330\,U^{2}.$$

To get the required answer, we must add all the entries of $$M$$. Because summation of matrix elements is a linear operation, we can add the totals contributed by $$I,\;U$$ and $$U^{2}$$ separately.

The sum of the elements of the identity matrix is clearly

$$\text{Sum}(I)=1+1+1=3.$$

The matrix $$U$$ is

$$U=\begin{bmatrix}0&1&1\\0&0&1\\0&0&0\end{bmatrix},$$

so

$$\text{Sum}(U)=0+1+1+0+0+1+0+0+0=3.$$

The matrix $$U^{2}$$ is

$$U^{2}=\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix},$$

thus

$$\text{Sum}(U^{2})=0+0+1+0+0+0+0+0+0=1.$$

Consequently, the total sum of all elements of $$M$$ equals

$$ \begin{aligned} \text{Sum}(M) &=20\cdot\text{Sum}(I)+210\cdot\text{Sum}(U)+1330\cdot\text{Sum}(U^{2})\\ &=20\cdot3+210\cdot3+1330\cdot1\\ &=60+630+1330\\ &=2020. \end{aligned} $$

So, the answer is $$2020$$.