if the product of n matrices `[(1,1),(0,1)][(1,2),(0,1)][(1,3),(0,1)]…[(1,n),(0,1)] is equal to the matrix [(1,378),(0,1)]` the value of n is equal to

Options

A. 26
B. 27
C. 377
D. 378

Answer

To solve the problem, we need to find the value of \( n \) such that the product of the matrices

\[ \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}, \ldots, \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} \]

equals

\[ \begin{pmatrix} 1 & 378 \\ 0 & 1 \end{pmatrix}. \]

### Step 1: Understand the Matrix Product

The product of two matrices of the form

\[ \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & a + b \\ 0 & 1 \end{pmatrix} \]

indicates that when we multiply these matrices, the resulting matrix will still be of the form

\[ \begin{pmatrix} 1 & c \\ 0 & 1 \end{pmatrix} \]

where \( c \) is the sum of the values \( a \) and \( b \).

### Step 2: Calculate the Product of Matrices

Now, we can generalize this to \( n \) matrices. The product of the first \( n \) matrices will yield:

\[ \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \cdots \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 + 2 + 3 + \ldots + n \\ 0 & 1 \end{pmatrix}. \]

### Step 3: Use the Formula for the Sum of the First n Natural Numbers

The sum of the first \( n \) natural numbers is given by the formula:

\[ S_n = \frac{n(n + 1)}{2}. \]

Thus, the product of the matrices results in:

\[ \begin{pmatrix} 1 & \frac{n(n + 1)}{2} \\ 0 & 1 \end{pmatrix}. \]

### Step 4: Set Up the Equation

We want this to equal

\[ \begin{pmatrix} 1 & 378 \\ 0 & 1 \end{pmatrix}. \]

This gives us the equation:

\[ \frac{n(n + 1)}{2} = 378. \]

### Step 5: Solve for n

Multiplying both sides by 2 gives:

\[ n(n + 1) = 756. \]

This can be rearranged to:

\[ n^2 + n - 756 = 0. \]

### Step 6: Factor the Quadratic Equation

To solve the quadratic equation \( n^2 + n - 756 = 0 \), we can factor it:

\[ (n - 27)(n + 28) = 0. \]

Setting each factor to zero gives:

\[ n - 27 = 0 \quad \text{or} \quad n + 28 = 0. \]

Thus, \( n = 27 \) or \( n = -28 \). Since \( n \) must be a positive integer, we have:

\[ n = 27. \]

### Conclusion

The value of \( n \) is \( 27 \).

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