Question:

Step 1: Define the general form of the matrices and calculate the product of two such matrices. Let the $k$-th matrix be denoted by $A_k$. From the problem statement, the general form of a matrix in the product sequence is: $$ A_k = \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix} $$ Now, let's find the product of two such matrices, $A_i$ and $A_j$: $$ A_i A_j = \begin{bmatrix} 1 & i \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & j \\ 0 & 1 \end{bmatrix} $$ Multiplying these matrices, we get: $$ A_i A_j = \begin{bmatrix} (1)(1)+(i)(0) & (1)(j)+(i)(1) \\ (0)(1)+(1)(0) & (0)(j)+(1)(1) \end{bmatrix} = \begin{bmatrix} 1 & i+j \\ 0 & 1 \end{bmatrix} $$ Step 2: Generalize the product of $n$ matrices. From the result in Step 1, we observe that when two matrices of this form are multiplied, their top-right elements add up. Extending this pattern, the product of $n$ matrices $A_1, A_2, \dots, A_n$ will result in a matrix where the top-right element is the sum of the top-right elements of the individual matrices. The product of the given $n$ matrices is: $$ \prod_{k=1}^n A_k = A_1 A_2 \dots A_n = \begin{bmatrix} 1 & 1+2+\dots+n \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & \sum_{k=1}^n k \\ 0 & 1 \end{bmatrix} $$ Using the formula for the sum of the first $n$ natural numbers, $\sum_{k=1}^n k = \frac{n(n+1)}{2}$, the product matrix can be written as: $$ \begin{bmatrix} 1 & \frac{n(n+1)}{2} \\ 0 & 1 \end{bmatrix} $$ Step 3: Equate the resulting product matrix to the given target matrix. The problem states that the product of the $n$ matrices is equal to the matrix $\begin{bmatrix} 1 & 378 \\ 0 & 1 \end{bmatrix}$. Equating our derived product matrix with the given target matrix: $$ \begin{bmatrix} 1 & \frac{n(n+1)}{2} \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 378 \\ 0 & 1 \end{bmatrix} $$ By comparing the corresponding elements of the two matrices, we can equate the top-right elements: $$ \frac{n(n+1)}{2} = 378 $$ Step 4: Solve the resulting quadratic equation for $n$. To solve for $n$, first, multiply both sides of the equation by 2: $$ n(n+1) = 378 \times 2 $$ $$ n^2 + n = 756 $$ Rearrange the terms to form a standard quadratic equation: $$ n^2 + n - 756 = 0 $$ We can factor this quadratic equation. We need two numbers whose product is $-756$ and whose sum is $1$. These numbers are $28$ and $-27$. $$ (n+28)(n-27) = 0 $$ This equation yields two possible values for $n$: $$ n = -28 \quad \text{or} \quad n = 27 $$ Since $n$ represents the number of matrices in a product, it must be a positive integer. Therefore, $n > 0$. Thus, we discard the negative solution $n = -28$. The valid value for $n$ is $27$. Step 5: State the final answer. The value of $n$ is 27. This corresponds to Option (B). The final answer is $\boxed{\text{27}}$.