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(B) Given $A = \begin{bmatrix} 1 & 1 & 1 \ 0 & 1 & 1 \ 0 & 0 & 1 \end{bmatrix}$.We calculate the powers of $A$:$A^2 = A \cdot A = \begin{bmatrix} 1

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$\begin{bmatrix} 1 & n & \frac{n(n+1)}{2} \ 0 & 1 & n \ 0 & 0 & 1 \end{bmatrix}$

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(B) Given $A = \begin{bmatrix} 1 & 1 & 1 \ 0 & 1 & 1 \ 0 & 0 & 1 \end{bmatrix}$.We calculate the powers of $A$:$A^2 = A \cdot A = \begin{bmatrix} 1 & 1 & 1 \ 0 & 1 & 1 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \ 0 & 1 & 1 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \ 0 & 1 & 2 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 & \frac{2(2+1)}{2} \ 0 & 1 & 2 \ 0 & 0 & 1 \end{bmatrix}$.$A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 2 & 3 \ 0 & 1 & 2 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \ 0 & 1 & 1 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 & 6 \ 0 & 1 & 3 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 & \frac{3(3+1)}{2} \ 0 & 1 & 3 \ 0 & 0 & 1 \end{bmatrix}$.By observing the pattern, for any positive integer $n$, we have:$A^n = \begin{bmatrix} 1 & n & \frac{n(n+1)}{2} \ 0 & 1 & n \ 0 & 0 & 1 \end{bmatrix}$.

Let $A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$. Then, for a positive integer $n$, $A^n$ is:

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