For each `n in N, n(n+1) (2n+1)` is divisible by

Options

A. (a) 15
B. (b) 8
C. (c) 7
D. (d) 6

Answer

To solve the problem, we need to show that the expression \( n(n+1)(2n+1) \) is divisible by a certain number for all natural numbers \( n \). We will check the divisibility by the options provided: 15, 8, 7, and 6.

### Step-by-Step Solution

1. **Define the expression**: Let \( P(n) = n(n+1)(2n+1) \). We need to analyze this expression for natural numbers \( n \).

2. **Substitute values for \( n \)**:
- Start with \( n = 1 \):
\[ P(1) = 1(1+1)(2 \cdot 1 + 1) = 1 \cdot 2 \cdot 3 = 6 \]
- Next, try \( n = 2 \):
\[ P(2) = 2(2+1)(2 \cdot 2 + 1) = 2 \cdot 3 \cdot 5 = 30 \]
- Now try \( n = 3 \):
\[ P(3) = 3(3+1)(2 \cdot 3 + 1) = 3 \cdot 4 \cdot 7 = 84 \]

3. **Check divisibility by 6**:
- We found \( P(1) = 6 \), \( P(2) = 30 \), and \( P(3) = 84 \).
- All these values are divisible by 6:
- \( 6 \div 6 = 1 \)
- \( 30 \div 6 = 5 \)
- \( 84 \div 6 = 14 \)

4. **General proof of divisibility by 6**:
- Note that \( n(n+1) \) is always even (since one of \( n \) or \( n+1 \) is even).
- Among any three consecutive integers \( n, n+1, 2n+1 \), at least one must be divisible by 3.
- Therefore, \( P(n) \) is divisible by both 2 and 3, which means it is divisible by \( 6 \).

5. **Conclusion**: Since \( P(n) \) is divisible by 6 for all natural numbers \( n \), we conclude that the answer is option D: 6.