1.1.7

$$\text{Let } A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{bmatrix},$$ so that we seek a formula for $$A^n$$. Observe $$A^2 = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{bmatrix},$$

$$A^3 = \begin{bmatrix} 1 & 3 & 6 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \\ \end{bmatrix},

A^4 = \begin{bmatrix} 1 & 4 & 10 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \\ \end{bmatrix}, \text{and }

A^5 = \begin{bmatrix} 1 & 5 & 15 \\ 0 & 1 & 5 \\ 0 & 0 & 1 \\ \end{bmatrix}.$$

We're now ready to make an educated guess at the formula in question. From the above computations, we suspect that $$A^n = \begin{bmatrix} 1 & n & \frac{n(n + 1)}{2} \\ 0 & 1 & n \\ 0 & 0 & 1 \\ \end{bmatrix}.$$

To prove that our guess is correct, we proceed by induction. Clearly our guess holds for $$n = 1$$. Next, we will show that if the formula for $$A^n$$ holds when $$n=p$$, then it also holds for $$n = p + 1$$:

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

\begin{bmatrix} 1 & p & \frac{p(p + 1)}{2} \\ 0 & 1 & p \\ 0 & 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} 1 & p + 1 & \frac{(p + 1)(p + 2)}{2} \\ 0 & 1 & p + 1 \\ 0 & 0 & 1 \\ \end{bmatrix}.$$

By our induction hypothesis, the formula for $$A^p$$ holds; therefore, the formula holds when $$n = p + 1$$ as well. By the Principle of Mathematical Induction, our formula holds for all natural numbers.