Let’s solve the problem step by step based on the image provided.

We aim to prove that the square of a number that is 1 more than a multiple of 3 is also 1 more than a multiple of 3. Specifically, the question asks us to expand and simplify $(3n + 1)^2$ and show that it meets this property.

Step 1: Expand $(3n + 1)^2$

Using the binomial expansion formula: $(a + b)^2 = a^2 + 2ab + b^2,$ we expand $(3n + 1)^2$: $(3n + 1)^2 = (3n)^2 + 2(3n)(1) + 1^2.$

Step 2: Simplify each term

$(3n)^2 = 9n^2, \quad 2(3n)(1) = 6n, \quad \text{and} \quad 1^2 = 1.$

Thus: $(3n + 1)^2 = 9n^2 + 6n + 1.$

Step 3: Factor out $3$ from the first two terms

We rewrite $9n^2 + 6n + 1$ as: $(3n + 1)^2 = 3(3n^2 + 2n) + 1.$

Step 4: Interpret the result

The term $3(3n^2 + 2n)$ is a multiple of 3. Adding 1 to it results in a number that is 1 more than a multiple of 3.

Final Conclusion

Thus, we have proven that the square of a number that is 1 more than a multiple of 3 is also 1 more than a multiple of 3.

Would you like further clarifications or a detailed explanation of any step? Let me know!

Here are 5 related questions for further exploration:

1. How would you prove this property for a number 2 more than a multiple of 3?
2. What happens when you square a number that is a multiple of 3?
3. How would this proof change if the number were of the form $3n - 1$?
4. Can we generalize this concept to other moduli, like numbers that are $1 \mod 5$?
5. What role does factoring play in modular arithmetic proofs like this?

Tip: When solving proofs involving multiples or modular arithmetic, always look for patterns that involve factoring or grouping terms to simplify your work.