To prove algebraically that the expression [tex]\((4n + 1)^2 - (2n - 1)\)[/tex] is an even number, let’s start by simplifying the expression step-by-step.
1. Expand the square term: First, we expand [tex]\((4n + 1)^2\)[/tex].
[tex]\[
(4n + 1)^2 = (4n)^2 + 2 \cdot 4n \cdot 1 + 1^2
\][/tex]
Simplifying this:
[tex]\[
(4n + 1)^2 = 16n^2 + 8n + 1
\][/tex]
2. Combine the expanded form with the rest of the expression: We then subtract [tex]\((2n - 1)\)[/tex] from the expanded form:
[tex]\[
(4n + 1)^2 - (2n - 1) = 16n^2 + 8n + 1 - (2n - 1)
\][/tex]
Distribute the negative sign and simplify:
[tex]\[
16n^2 + 8n + 1 - 2n + 1 = 16n^2 + 6n + 2
\][/tex]
3. Determine if the simplified expression is even: An even number can be written in the form [tex]\(2k\)[/tex] where [tex]\(k\)[/tex] is an integer. Let’s check if [tex]\(16n^2 + 6n + 2\)[/tex] can be expressed in such a way.
To see this more clearly, we can factor out a 2 from the simplified expression:
[tex]\[
16n^2 + 6n + 2 = 2(8n^2 + 3n + 1)
\][/tex]
Since [tex]\(8n^2 + 3n + 1\)[/tex] is an integer for all integer values of [tex]\(n\)[/tex], the entire expression [tex]\(2(8n^2 + 3n + 1)\)[/tex] is clearly a multiple of 2.
Therefore, [tex]\(16n^2 + 6n + 2\)[/tex] is even and we've shown algebraically that [tex]\((4n + 1)^2 - (2n - 1)\)[/tex] is indeed an even number.
