Let $$A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{bmatrix}$$ and $$B = A^{20}$$. Then the sum of the elements of the first column of B is:

We have the lower-triangular matrix

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

and we want the sum of the three numbers that appear in the first column of $$B=A^{20}\,.$$ Let us denote, for every positive integer $$n,$$

$$A^{n}=\begin{bmatrix}a_{11}^{(n)}&a_{12}^{(n)}&a_{13}^{(n)}\\[2pt] a_{21}^{(n)}&a_{22}^{(n)}&a_{23}^{(n)}\\[2pt] a_{31}^{(n)}&a_{32}^{(n)}&a_{33}^{(n)}\end{bmatrix}.$$

The required sum is therefore

$$S_{n}=a_{11}^{(n)}+a_{21}^{(n)}+a_{31}^{(n)}.$$

We now build up the powers step by step so that every algebraic detail is visible and a pattern can be recognised.

Power 1. We clearly have

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

so

$$\bigl(a_{11}^{(1)},\,a_{21}^{(1)},\,a_{31}^{(1)}\bigr)=(1,1,1)$$

and therefore

$$S_{1}=1+1+1=3.$$

Power 2. Using plain matrix multiplication, and writing every scalar product explicitly, we have

$$A^{2}=A\!\cdot\!A=\begin{bmatrix} 1\cdot1+0\cdot1+0\cdot1 & 1\cdot0+0\cdot1+0\cdot1 & 1\cdot0+0\cdot0+0\cdot1\\[4pt] 1\cdot1+1\cdot1+0\cdot1 & 1\cdot0+1\cdot1+0\cdot1 & 1\cdot0+1\cdot0+0\cdot1\\[4pt] 1\cdot1+1\cdot1+1\cdot1 & 1\cdot0+1\cdot1+1\cdot1 & 1\cdot0+1\cdot0+1\cdot1 \end{bmatrix}$$

which simplifies to

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

Thus

$$\bigl(a_{11}^{(2)},\,a_{21}^{(2)},\,a_{31}^{(2)}\bigr)=(1,2,3)$$

and

$$S_{2}=1+2+3=6.$$

Power 3. Multiplying $$A^{2}$$ by $$A$$ again, while displaying every dot product,

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

gives

$$\begin{aligned} A^{3}&=\begin{bmatrix} 1\cdot1+0\cdot1+0\cdot1 & 1\cdot0+0\cdot1+0\cdot1 & 1\cdot0+0\cdot0+0\cdot1\\[4pt] 2\cdot1+1\cdot1+0\cdot1 & 2\cdot0+1\cdot1+0\cdot1 & 2\cdot0+1\cdot0+0\cdot1\\[4pt] 3\cdot1+2\cdot1+1\cdot1 & 3\cdot0+2\cdot1+1\cdot1 & 3\cdot0+2\cdot0+1\cdot1 \end{bmatrix}\\[6pt] &=\begin{bmatrix}1&0&0\\3&1&0\\6&3&1\end{bmatrix}. \end{aligned}$$

Hence

$$\bigl(a_{11}^{(3)},\,a_{21}^{(3)},\,a_{31}^{(3)}\bigr)=(1,3,6)$$

and

$$S_{3}=1+3+6=10.$$

Writing the sums that we have found so far, we notice the sequence

$$S_{1}=3,\qquad S_{2}=6,\qquad S_{3}=10.$$

The successive differences are $$3$$ and $$4,$$ then $$4$$ and $$?$$ in the next step, so it looks like the pattern is the familiar triangular-number pattern. Indeed, the first three terms can be matched exactly by the well-known formula for the $$k$$-th triangular number, namely

$$T_{k}=\frac{k(k+1)}{2}.$$

Re-indexing appropriately, we suspect that

$$S_{n}=\frac{(n+1)(n+2)}{2}=\binom{n+2}{2}.$$

Formal proof of the pattern by mathematical induction.

Base case $$n=1$$ has already been calculated: $$S_{1}=3=\binom{3}{2}.$$ Now assume for some $$n=k$$ that

$$S_{k}=a_{11}^{(k)}+a_{21}^{(k)}+a_{31}^{(k)}=\binom{k+2}{2}.$$

To pass from $$A^{k}$$ to $$A^{k+1},$$ we right-multiply by $$A.$$ Concerning only the first column, we use the fact that the first column of $$A$$ itself is

$$\begin{bmatrix}1\\1\\1\end{bmatrix},$$

so

$$\begin{bmatrix}a_{11}^{(k+1)}\\ a_{21}^{(k+1)}\\ a_{31}^{(k+1)}\end{bmatrix}=A^{k}\begin{bmatrix}1\\1\\1\end{bmatrix}= \begin{bmatrix} a_{11}^{(k)}\cdot1+a_{12}^{(k)}\cdot1+a_{13}^{(k)}\cdot1\\[4pt] a_{21}^{(k)}\cdot1+a_{22}^{(k)}\cdot1+a_{23}^{(k)}\cdot1\\[4pt] a_{31}^{(k)}\cdot1+a_{32}^{(k)}\cdot1+a_{33}^{(k)}\cdot1 \end{bmatrix}.$$

That is,

$$\begin{aligned} a_{11}^{(k+1)}&=a_{11}^{(k)}+a_{12}^{(k)}+a_{13}^{(k)},\\ a_{21}^{(k+1)}&=a_{21}^{(k)}+a_{22}^{(k)}+a_{23}^{(k)},\\ a_{31}^{(k+1)}&=a_{31}^{(k)}+a_{32}^{(k)}+a_{33}^{(k)}. \end{aligned}$$

Adding these three equalities together we get

$$S_{k+1}=S_{k}+ \bigl(a_{12}^{(k)}+a_{22}^{(k)}+a_{32}^{(k)}\bigr)+ \bigl(a_{13}^{(k)}+a_{23}^{(k)}+a_{33}^{(k)}\bigr).$$

But $$A^{k}$$ is lower triangular with ones on the diagonal, so $$a_{12}^{(k)}=0$$ and $$a_{13}^{(k)}=a_{23}^{(k)}=0.$$ Therefore

$$S_{k+1}=S_{k}+a_{22}^{(k)}+a_{32}^{(k)}+a_{33}^{(k)}.$$

Since every diagonal entry of any power of a unit lower-triangular matrix is $$1,$$ we know $$a_{22}^{(k)}=a_{33}^{(k)}=1.$$ Furthermore, the (3,2) entry is precisely $$k,$$ because we can check directly for $$k=1,2,3$$ and see by induction that one unit is added at each new multiplication. Hence

$$S_{k+1}=S_{k}+1+k+1=S_{k}+k+2.$$

Using the induction hypothesis $$S_{k}=\dfrac{(k+1)(k+2)}{2},$$ we get

$$\begin{aligned} S_{k+1}&=\frac{(k+1)(k+2)}{2}+k+2\\[4pt] &=\frac{(k+1)(k+2)+2(k+2)}{2}\\[4pt] &=\frac{(k+2)(k+3)}{2}\\[4pt] &=\binom{(k+1)+2}{2}. \end{aligned}$$

Thus the induction is complete and the formula

$$S_{n}=\frac{(n+1)(n+2)}{2}=\binom{n+2}{2}$$

holds for every positive integer $$n.$$

Evaluating for $$n=20.$$

Substituting $$n=20$$ in the proven expression, we obtain

$$S_{20}=\frac{(20+1)(20+2)}{2}=\frac{21\cdot22}{2}=231.$$

Therefore, the sum of the elements of the first column of $$B=A^{20}$$ equals $$231.$$

Hence, the correct answer is Option D.