Question

If n be any integer, then \(n(n+1)(2n+1)\) is:

• A. odd number
• B. integral multiple of 6
• C. perfect square
• D. does not necessarily have any of the foregoing proof

Hint:

Product of three consecutive integers is divisible by 6.

Answer 1

The Correct Option is B

To determine whether the expression \(n(n+1)(2n+1)\) is an integral multiple of 6 for any integer \(n\), we need to analyze the expression step by step.

The expression \(n(n+1)(2n+1)\) consists of three consecutive terms:

• \(n\)
• \((n+1)\)
• \((2n+1)\)

This expression is essentially a product of three numbers: \(n\), \(n+1\), and \(2n+1\).

1. Every integer \(n\) will ensure that either \(n\) or \(n+1\) is even. This means the product \(n(n+1)\) is always even, and hence \(n(n+1)(2n+1)\) is even.
2. For the product to be a multiple of 6, it should also be divisible by 3. For any integer \(n\), at least one of the three terms in the expression \(n(n+1)(2n+1)\) must be divisible by 3, due to the nature of successive integers. Specifically:
   • If \(n\) is divisible by 3, then the entire expression is divisible by 3.
   • If \(n \equiv 1 \pmod{3}\), then \(n + 1\) is divisible by 3.
   • If \(n \equiv 2 \pmod{3}\), then \(2n + 1\) becomes \(4k + 5\equiv 0 \pmod{3}\).

Therefore, by combining both of these divisibility checks, the expression \(n(n+1)(2n+1)\) will always have factors of both 2 and 3, making it a multiple of 6.

Hence, the expression \(n(n+1)(2n+1)\) is always an integral multiple of 6 for any integer \(n\).