Certainly! Let's work through the proof step-by-step.
Given expression: [tex]\((3n + 1)^2\)[/tex]
We need to expand and simplify this expression to prove the statement.
### Step 1: Expand the expression
Let's expand the square:
[tex]\[
(3n + 1)^2 = (3n + 1)(3n + 1)
\][/tex]
Applying the distributive property (FOIL method):
[tex]\[
(3n + 1)(3n + 1) = 3n \cdot 3n + 3n \cdot 1 + 1 \cdot 3n + 1 \cdot 1
\][/tex]
[tex]\[
= 9n^2 + 3n + 3n + 1
\][/tex]
[tex]\[
= 9n^2 + 6n + 1
\][/tex]
### Step 2: Identify the coefficients
From the expanded expression [tex]\(9n^2 + 6n + 1\)[/tex]:
[tex]\[
\begin{aligned}
n^2 \text{ term: } & 9n^2, \quad \text{(coefficient is 9)} \\
n \text{ term: } & 6n, \quad \text{(coefficient is 6)} \\
\text{constant term: } & 1
\end{aligned}
\][/tex]
Thus, we can write:
[tex]\[
(3n + 1)^2 = 9n^2 + 6n + 1
\][/tex]
### Step 3: Factor out a multiple of 3 and simplify
We notice that [tex]\(9n^2 + 6n\)[/tex] is a multiple of 3:
[tex]\[
9n^2 + 6n = 3(3n^2 + 2n)
\][/tex]
So, we can rewrite the expression as:
[tex]\[
9n^2 + 6n + 1 = 3(3n^2 + 2n) + 1
\][/tex]
This shows that [tex]\(9n^2 + 6n\)[/tex] is a multiple of 3, and when we add 1, we get a number that is one more than a multiple of 3.
### Conclusion
[tex]\[
\begin{aligned}
(3n + 1)^2 &= 9n^2 + 6n + 1 \\
&= 3(3n^2 + 2n) + 1
\end{aligned}
\][/tex]
This confirms that [tex]\((3n + 1)^2\)[/tex] is indeed one more than a multiple of 3.
Therefore, we have proven that the square of a number that is one more than a multiple of 3 is also one more than a multiple of 3.
