The problem asks to show that $n(n+1)(2n+1)$ is a multiple of $6$ for every natural number $n$. Let’s break it down step by step:

Proof:

1. Expression Analysis:

   • The given expression is $n(n+1)(2n+1)$.
   • Here, $n$, $n+1$, and $2n+1$ are consecutive integers or terms closely related.

2. Divisibility by 2:

   • Among $n$ and $n+1$, one is always even because consecutive integers include one even number.
   • Therefore, $n(n+1)(2n+1)$ is always divisible by $2$.

3. Divisibility by 3:

   • Among $n$, $n+1$, and $2n+1$, at least one is divisible by $3$:
     • If $n$ is divisible by $3$, the product is divisible by $3$.
     • If $n \mod 3 = 1$, then $n+1$ is divisible by $3$.
     • If $n \mod 3 = 2$, then $2n+1 = 2(2) + 1 = 5 \mod 3 = 0$.

   Therefore, $n(n+1)(2n+1)$ is always divisible by $3$.

4. Divisibility by 6:

   • Since the expression is divisible by both $2$ and $3$, it is divisible by $6$.

Conclusion:

For any natural number $n$, $n(n+1)(2n+1)$ is divisible by $6$.

Would you like me to explain any step in more detail? Here are some follow-up questions to extend your understanding:

1. What is the significance of testing divisibility by both 2 and 3 to prove divisibility by 6?
2. How can this method of proof be generalized for proving divisibility by other composite numbers?
3. Can you test this with specific values of $n$ to verify the result?
4. Why does $2n+1$ necessarily contribute to divisibility in this problem?
5. How does this problem demonstrate properties of consecutive integers?

Tip: Always check divisibility rules step-by-step when proving divisibility for composite numbers like $6$.