If `[{:(1, 1), (0,1):}]*[{:(1, 2), (0,1):}]*[{:(1, 3), (0,1):}]cdotcdotcdot[{:(1, n-1), (0,1):}] = [{:(1, 78), (0,1):}]`, then the inverse of `[{:(1, n), (0,1):}]` is

Options

A. `[{:(1, 0), (12, 1):}]`
B. `[{:(1, -13), (0, 1):}]`
C. `[{:(1, 0), (13, 1):}]`
D. `[{:(1, -12), (0, 1):}]`

Explanation

To solve the given problem, we need to find the inverse of the matrix \(\begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}\) given that the product of a series of matrices equals \(\begin{pmatrix} 1 & 78 \\ 0 & 1 \end{pmatrix}\).

### Step-by-step Solution:

1. **Understanding the Matrix Product**:
We are given a product of matrices of the form:
\[ \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \cdots \begin{pmatrix} 1 & n-1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 78 \\ 0 & 1 \end{pmatrix} \]

2. **Calculating the Product**:
The product of two matrices of the form \(\begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}\) and \(\begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix}\) results in:
\[ \begin{pmatrix} 1 & a+b \\ 0 & 1 \end{pmatrix} \]
Therefore, the product of all matrices from \(1\) to \(n-1\) will yield:
\[ \begin{pmatrix} 1 & 1 + 2 + 3 + \ldots + (n-1) \\ 0 & 1 \end{pmatrix} \]

3. **Using the Sum of Natural Numbers**:
The sum of the first \(n-1\) natural numbers is given by:
\[ S = \frac{(n-1)n}{2} \]
Thus, we have:
\[ \begin{pmatrix} 1 & \frac{(n-1)n}{2} \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 78 \\ 0 & 1 \end{pmatrix} \]

4. **Setting Up the Equation**:
From the equality of the matrices, we can equate the upper right elements:
\[ \frac{(n-1)n}{2} = 78 \]
Multiplying both sides by \(2\):
\[ (n-1)n = 156 \]

5. **Forming a Quadratic Equation**:
Rearranging gives us:
\[ n^2 - n - 156 = 0 \]

6. **Factoring the Quadratic**:
We can factor this as:
\[ n^2 - 13n + 12n - 156 = 0 \]
This factors to:
\[ (n - 13)(n + 12) = 0 \]
Thus, \(n = 13\) or \(n = -12\). Since \(n\) must be a positive integer, we take \(n = 13\).

7. **Finding the Matrix**:
The matrix we are interested in is:
\[ A = \begin{pmatrix} 1 & 13 \\ 0 & 1 \end{pmatrix} \]

8. **Calculating the Inverse**:
The inverse of a matrix of the form \(\begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}\) is given by:
\[ A^{-1} = \begin{pmatrix} 1 & -a \\ 0 & 1 \end{pmatrix} \]
Therefore, the inverse is:
\[ A^{-1} = \begin{pmatrix} 1 & -13 \\ 0 & 1 \end{pmatrix} \]

### Final Answer:
The inverse of \(\begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}\) when \(n = 13\) is:
\[ \begin{pmatrix} 1 & -13 \\ 0 & 1 \end{pmatrix} \]