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

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

is equal to

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

### Step 1: Understand the Matrix Multiplication
The product of two matrices of the form

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

results in

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

### Step 2: Generalize the Product
When we multiply \( n \) such matrices, we can see that the resulting matrix will be

\[ \begin{pmatrix} 1 & (1 + 2 + 3 + \ldots + n) \\ 0 & 1 \end{pmatrix}. \]

### 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

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

### Step 4: Set Up the Equation
We need this sum to equal 378:

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

### Step 5: Solve for \( n \)
Multiply both sides by 2:

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

### Step 6: Rearrange into a Quadratic Equation
This gives us the quadratic equation:

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

### Step 7: Factor the Quadratic Equation
To factor, we look for two numbers that multiply to -756 and add to 1. The factors are 28 and -27:

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

### Step 8: Solve for \( n \)
Setting each factor to zero gives:

1. \( n + 28 = 0 \) → \( n = -28 \) (not a valid solution since \( n \) must be a natural number)
2. \( n - 27 = 0 \) → \( n = 27 \).

### Conclusion
Thus, the value of \( n \) is

\[ \boxed{27}. \]