Certainly! Let's go through the problem step-by-step to prove that the square of a number that is one more than a multiple of 3 is also one more than a multiple of 3.
We start with the expression [tex]\((3n + 1)^2\)[/tex].
1. Expand the expression:
[tex]\[
(3n + 1)^2 = (3n + 1)(3n + 1)
\][/tex]
Use the distributive property (also known as FOIL in this specific case):
[tex]\[
(3n + 1)(3n + 1) = (3n)(3n) + (3n)(1) + (1)(3n) + (1)(1)
\][/tex]
2. Calculate each term separately:
[tex]\[
(3n)(3n) = 9n^2
\][/tex]
[tex]\[
(3n)(1) = 3n
\][/tex]
[tex]\[
(1)(3n) = 3n
\][/tex]
[tex]\[
(1)(1) = 1
\][/tex]
3. Add all the terms together:
[tex]\[
9n^2 + 3n + 3n + 1
\][/tex]
Combine like terms:
[tex]\[
9n^2 + 6n + 1
\][/tex]
We now have the expanded expression:
[tex]\[
(3n + 1)^2 = 9n^2 + 6n + 1
\][/tex]
4. Analyze the result:
Notice that [tex]\(9n^2 + 6n\)[/tex] is a multiple of 3 because both [tex]\(9n^2\)[/tex] and [tex]\(6n\)[/tex] are individually multiples of 3.
Indeed, 9 and 6 are multiples of 3:
[tex]\[
9n^2 = 3 \cdot 3n^2 \quad \text{and} \quad 6n = 3 \cdot 2n
\][/tex]
5. Express [tex]\(9n^2 + 6n + 1\)[/tex] as one more than a multiple of 3:
Factor the expression to show it clearly:
[tex]\[
9n^2 + 6n + 1 = 3(3n^2 + 2n) + 1
\][/tex]
This shows that the term [tex]\(9n^2 + 6n\)[/tex] is a multiple of 3, and adding 1 makes the whole expression [tex]\(9n^2 + 6n + 1\)[/tex] one more than a multiple of 3.
Therefore, [tex]\((3n + 1)^2\)[/tex] is indeed one more than a multiple of 3.
In conclusion:
[tex]\[
(3n + 1)^2 = 9n^2 + 6n + 1
\][/tex]
is clearly shown to be:
[tex]\[
1 \text{ more than a multiple of } 3
\][/tex]
