What is the `15^(th)` term of the sequence defined by:
`a_(n)=(n-1)(2-n)(3+n)?`

Options

A. -3276
B. 3298
C. -2764
D. -9874

Answer

To find the 15th term of the sequence defined by the formula \( a_n = (n-1)(2-n)(3+n) \), we will follow these steps:

### Step 1: Substitute \( n = 15 \) into the formula
We need to find \( a_{15} \). So we substitute \( n = 15 \) into the formula:

\[ a_{15} = (15-1)(2-15)(3+15) \]

### Step 2: Simplify each part of the expression
Now, we simplify each term in the expression:

- \( 15 - 1 = 14 \)
- \( 2 - 15 = -13 \)
- \( 3 + 15 = 18 \)

So, we can rewrite the expression as:

\[ a_{15} = 14 \cdot (-13) \cdot 18 \]

### Step 3: Calculate the product
Now, we will calculate the product step by step:

1. First, calculate \( 14 \cdot (-13) \):
\[ 14 \cdot (-13) = -182 \]

2. Next, multiply this result by \( 18 \):
\[ -182 \cdot 18 \]

To calculate \( -182 \cdot 18 \), we can do the multiplication:
\[ 182 \cdot 18 = 3276 \]
Since we have a negative sign, we get:
\[ -182 \cdot 18 = -3276 \]

### Final Result
Thus, the 15th term of the sequence is:
\[ \boxed{-3276} \]

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