Let the `n^(th)` term of a series be given by `t_n = (n^2 -n-2)/(n^2+3n),n leq 3`. Then product `t_3 t_4 .....t_50` is equal to

Options

A. `(1)/(5^(2).7.13.53)`
B. `(1)/(5.7^(2).12.53)`
C. `(1)/(5^(2).7.12.51)`
D. `(1)/(5.7^(2).13.53)`

Answer

To solve the problem, we need to find the product \( t_3 t_4 \ldots t_{50} \) where the \( n^{th} \) term is given by

\[ t_n = \frac{n^2 - n - 2}{n^2 + 3n} \]

### Step 1: Factor the numerator and denominator

First, we will factor both the numerator and the denominator.

**Numerator:**
\[ n^2 - n - 2 = (n - 2)(n + 1) \]

**Denominator:**
\[ n^2 + 3n = n(n + 3) \]

Thus, we can rewrite \( t_n \) as:

\[ t_n = \frac{(n - 2)(n + 1)}{n(n + 3)} \]

### Step 2: Write the product \( t_3 t_4 \ldots t_{50} \)

Now we can express the product \( t_3 t_4 \ldots t_{50} \):

\[ t_3 t_4 \ldots t_{50} = \prod_{n=3}^{50} \frac{(n - 2)(n + 1)}{n(n + 3)} \]

This can be separated into two products:

\[ = \frac{\prod_{n=3}^{50} (n - 2) \prod_{n=3}^{50} (n + 1)}{\prod_{n=3}^{50} n \prod_{n=3}^{50} (n + 3)} \]

### Step 3: Evaluate each product

1. **Evaluate \( \prod_{n=3}^{50} (n - 2) \)**:
- This is \( 1 \times 2 \times 3 \times \ldots \times 48 \) which is \( 48! \).

2. **Evaluate \( \prod_{n=3}^{50} (n + 1) \)**:
- This is \( 4 \times 5 \times 6 \times \ldots \times 51 \) which can be written as \( \frac{51!}{3!} \).

3. **Evaluate \( \prod_{n=3}^{50} n \)**:
- This is \( 3 \times 4 \times 5 \times \ldots \times 50 \) which can be written as \( \frac{50!}{2!} \).

4. **Evaluate \( \prod_{n=3}^{50} (n + 3) \)**:
- This is \( 6 \times 7 \times 8 \times \ldots \times 53 \) which can be written as \( \frac{53!}{5!} \).

### Step 4: Combine the products

Now we can substitute these products back into our equation:

\[ t_3 t_4 \ldots t_{50} = \frac{48! \cdot \frac{51!}{3!}}{\frac{50!}{2!} \cdot \frac{53!}{5!}} \]

### Step 5: Simplify the expression

This simplifies to:

\[ = \frac{48! \cdot 51! \cdot 2! \cdot 5!}{3! \cdot 50! \cdot 53!} \]

### Step 6: Calculate the final value

Now we can simplify further, noting that \( 50! = 50 \cdot 49 \cdot 48! \):

\[ = \frac{48! \cdot 51! \cdot 2! \cdot 5!}{3! \cdot 50 \cdot 49 \cdot 48! \cdot 53!} \]

After canceling \( 48! \):

\[ = \frac{51! \cdot 2! \cdot 5!}{3! \cdot 50 \cdot 49 \cdot 53!} \]

### Final Result

After calculating the factorials and simplifying, we find that the final product \( t_3 t_4 \ldots t_{50} \) equals:

\[ \frac{1}{5 \cdot 7^2 \cdot 13 \cdot 53} \]