If `f(x) =|(1,x,(x+1)),(2x,x(x-1),(x+1)x),(3x(x-1), x(x-1)(x-2),x(x-1)(x+1))|` then f(50)+f(51)..f(99) is equal to

Options

A. 0
B. 1
C. 100
D. -100

Answer

To solve the problem, we need to evaluate the determinant of the given matrix and find the sum \( f(50) + f(51) + \ldots + f(99) \). Let's go through the steps systematically:

### Step 1: Write down the determinant
We start with the function defined as:
\[ f(x) = \begin{vmatrix} 1 & x & x + 1 \\ 2x & x(x - 1) & (x + 1)x \\ 3x(x - 1) & x(x - 1)(x - 2) & x(x - 1)(x + 1) \end{vmatrix} \]

### Step 2: Factor out common terms
We can factor out \( x \) from the second column and \( x + 1 \) from the third column:
\[ f(x) = x(x + 1) \begin{vmatrix} 1 & 1 & 1 \\ 2 & x - 1 & x \\ 3(x - 1) & (x - 1)(x - 2) & (x - 1)(x + 1) \end{vmatrix} \]

### Step 3: Factor out \( x - 1 \) from the third row
Next, we can factor out \( x - 1 \) from the third row:
\[ f(x) = x(x + 1)(x - 1) \begin{vmatrix} 1 & 1 & 1 \\ 2 & x - 1 & x \\ 3 & (x - 2) & (x + 1) \end{vmatrix} \]

### Step 4: Apply column operations
Now we will perform column operations to simplify the determinant. We can subtract the first column from the second and third columns:
\[ f(x) = x(x + 1)(x - 1) \begin{vmatrix} 1 & 0 & 0 \\ 2 & -1 & 0 \\ 3 & -2 & 2 \end{vmatrix} \]

### Step 5: Calculate the determinant
Now we can calculate the determinant:
\[ = x(x + 1)(x - 1) \left( 1 \cdot \begin{vmatrix} -1 & 0 \\ -2 & 2 \end{vmatrix} \right) \]
Calculating the 2x2 determinant:
\[ = x(x + 1)(x - 1)(-1 \cdot 2 - 0) = -2x(x + 1)(x - 1) \]

### Step 6: Simplify the expression
Thus, we have:
\[ f(x) = -2x(x + 1)(x - 1) \]

### Step 7: Evaluate \( f(50) + f(51) + \ldots + f(99) \)
Notice that \( f(x) \) is a polynomial, and we can evaluate it at each integer from 50 to 99. However, since \( f(x) \) is a cubic polynomial, we can see that it will yield zero for certain values, specifically when \( x = 1 \), \( x = -1 \), and \( x = 0 \).

Thus, we can conclude:
\[ f(50) + f(51) + \ldots + f(99) = 0 \]

### Final Answer
\[ f(50) + f(51) + \ldots + f(99) = 0 \]