Certainly! Let's break down the solution step by step to prove that the square of a number that is two more than a multiple of 3 is one more than a multiple of 3.
Given a number of the form [tex]\(3n + 2\)[/tex], where [tex]\(n\)[/tex] is an integer, we want to find the square of this number and show that it's one more than a multiple of 3.
Step 1: Expand the square
We start by squaring the expression:
[tex]\[
(3n + 2)^2
\][/tex]
Step 2: Apply the algebraic expansion formula
Expanding using the formula for the square of a binomial [tex]\((a + b)^2 = a^2 + 2ab + b^2\)[/tex]:
[tex]\[
(3n + 2)^2 = (3n)^2 + 2(3n)(2) + (2)^2
\][/tex]
Step 3: Compute each term
Calculate each term separately:
[tex]\[
(3n)^2 = 9n^2
\][/tex]
[tex]\[
2(3n)(2) = 12n
\][/tex]
[tex]\[
(2)^2 = 4
\][/tex]
Step 4: Combine the terms
Add these terms together:
[tex]\[
(3n + 2)^2 = 9n^2 + 12n + 4
\][/tex]
Step 5: Group and simplify
Now let's rewrite this as:
[tex]\[
(3n + 2)^2 = 9n^2 + 12n + 3 + 1
\][/tex]
We recognize that [tex]\(9n^2 + 12n + 3\)[/tex] is a multiple of 3 because each term [tex]\(9n^2\)[/tex], [tex]\(12n\)[/tex] and 3 are individually multiples of 3. Let's factor out the 3:
[tex]\[
9n^2 + 12n + 3 = 3(3n^2 + 4n + 1)
\][/tex]
Hence:
[tex]\[
(3n + 2)^2 = 3(3n^2 + 4n + 1) + 1
\][/tex]
Step 6: Conclude
We have shown that:
[tex]\[
(3n + 2)^2\ =3(3n^2 + 4n + 1) + 1
\][/tex]
Which explicitly indicates that [tex]\((3n + 2)^2\)[/tex] is one more than a multiple of 3.
[tex]\[
= 1 \text{ more than a multiple of } 3
\][/tex]
Thus, we have proven that the square of a number that is two more than a multiple of 3 is one more than a multiple of 3.
