For all integral values of n, the largest number that exactly divides each number of the sequence `(n-1) n(n+1) , n(n+1) (n+2), (n+1) (n+2) (n+3)`…..is

Options

A. 12
B. 6
C. 3
D. 2

Answer

To find the largest number that exactly divides each number of the sequence given by \((n-1)n(n+1)\), \(n(n+1)(n+2)\), and \((n+1)(n+2)(n+3)\), we can analyze the expressions step by step.

### Step 1: Analyze the expressions
The three expressions in the sequence are:
1. \( (n-1)n(n+1) \)
2. \( n(n+1)(n+2) \)
3. \( (n+1)(n+2)(n+3) \)

### Step 2: Factor each expression
Let's factor each expression to identify common factors.
1. **First Expression**:
\[ (n-1)n(n+1) = n^3 - n \]
2. **Second Expression**:
\[ n(n+1)(n+2) = n(n^2 + 3n + 2) = n^3 + 3n^2 + 2n \]
3. **Third Expression**:
\[ (n+1)(n+2)(n+3) = (n^2 + 3n + 2)(n + 3) = n^3 + 6n^2 + 11n + 6 \]

### Step 3: Evaluate the expressions for specific values of \(n\)
To find the largest common divisor, we can evaluate the expressions for small integral values of \(n\).

- **For \(n = 1\)**:
- \( (1-1) \cdot 1 \cdot (1+1) = 0 \)
- \( 1 \cdot (1+1) \cdot (1+2) = 1 \cdot 2 \cdot 3 = 6 \)
- \( (1+1) \cdot (1+2) \cdot (1+3) = 2 \cdot 3 \cdot 4 = 24 \)

- **For \(n = 2\)**:
- \( (2-1) \cdot 2 \cdot (2+1) = 1 \cdot 2 \cdot 3 = 6 \)
- \( 2 \cdot (2+1) \cdot (2+2) = 2 \cdot 3 \cdot 4 = 24 \)
- \( (2+1) \cdot (2+2) \cdot (2+3) = 3 \cdot 4 \cdot 5 = 60 \)

- **For \(n = 3\)**:
- \( (3-1) \cdot 3 \cdot (3+1) = 2 \cdot 3 \cdot 4 = 24 \)
- \( 3 \cdot (3+1) \cdot (3+2) = 3 \cdot 4 \cdot 5 = 60 \)
- \( (3+1) \cdot (3+2) \cdot (3+3) = 4 \cdot 5 \cdot 6 = 120 \)

### Step 4: Identify the common divisor
From our calculations:
- For \(n = 1\): The values are \(0, 6, 24\) (ignore 0)
- For \(n = 2\): The values are \(6, 24, 60\)
- For \(n = 3\): The values are \(24, 60, 120\)

The common divisor across these values is \(6\).

### Conclusion
The largest number that exactly divides each number of the sequence for all integral values of \(n\) is **6**.