show that `n(n^2 -1)`, is divisible by 24 if n is an odd positive number.

Options

A. 6
B. 16
C. 36
D. 24

Answer

To show that \( n(n^2 - 1) \) is divisible by 24 for any odd positive integer \( n \), we will use mathematical induction.

### Step 1: Express \( n \) as an odd integer
An odd positive integer can be expressed in the form:
\[ n = 2k + 1 \]
where \( k \) is a non-negative integer.

### Step 2: Substitute \( n \) into the expression
Now, we substitute \( n \) into the expression \( n(n^2 - 1) \):
\[ n(n^2 - 1) = (2k + 1)((2k + 1)^2 - 1) \]
Calculating \( (2k + 1)^2 - 1 \):
\[ (2k + 1)^2 = 4k^2 + 4k + 1 \]
Thus,
\[ (2k + 1)^2 - 1 = 4k^2 + 4k + 1 - 1 = 4k^2 + 4k \]
Now substituting back, we have:
\[ n(n^2 - 1) = (2k + 1)(4k^2 + 4k) \]

### Step 3: Factor the expression
Factoring out \( 4k \) from \( 4k^2 + 4k \):
\[ n(n^2 - 1) = (2k + 1)(4k(k + 1)) \]

### Step 4: Analyze divisibility by 8
Notice that \( 4k(k + 1) \) is always divisible by 8 because:
- \( k(k + 1) \) is the product of two consecutive integers, which is always even.
- Therefore, \( 4k(k + 1) \) is divisible by \( 4 \times 2 = 8 \).

### Step 5: Analyze divisibility by 3
Next, we need to check if \( n(n^2 - 1) \) is divisible by 3. Since \( n \) is odd, \( n \equiv 1 \) or \( n \equiv 2 \mod 3 \):
1. If \( n \equiv 1 \mod 3 \):
\[ n^2 - 1 \equiv 1^2 - 1 \equiv 0 \mod 3 \]
2. If \( n \equiv 2 \mod 3 \):
\[ n^2 - 1 \equiv 2^2 - 1 \equiv 4 - 1 \equiv 0 \mod 3 \]

In both cases, \( n^2 - 1 \) is divisible by 3.

### Step 6: Combine results
Since \( n(n^2 - 1) \) is divisible by both 8 and 3, it is also divisible by:
\[ \text{lcm}(8, 3) = 24 \]

### Conclusion
Thus, we conclude that \( n(n^2 - 1) \) is divisible by 24 for any odd positive integer \( n \).

---