To prove that the square of the sum of two consecutive integers is odd, let's work through a detailed, step-by-step solution.
First, consider two consecutive integers [tex]\( n \)[/tex] and [tex]\( n+1 \)[/tex]. We need to prove that the square of their sum is odd.
1. Expression for the sum of two consecutive integers:
[tex]\[
n + (n + 1) = 2n + 1
\][/tex]
2. Square the sum:
[tex]\[
(2n + 1)^2
\][/tex]
3. Expand the squared expression:
[tex]\[
(2n + 1)^2 = (2n)^2 + 2 \cdot (2n) \cdot 1 + 1^2
\][/tex]
[tex]\[
(2n + 1)^2 = 4n^2 + 4n + 1
\][/tex]
4. Simplify the expanded expression:
[tex]\[
4n^2 + 4n + 1
\][/tex]
5. Extract coefficients:
Identifying the coefficients, we have:
- The coefficient of [tex]\( n^2 \)[/tex] is [tex]\( 4 \)[/tex].
- The coefficient of [tex]\( n \)[/tex] is [tex]\( 4 \)[/tex].
- The constant term is [tex]\( 1 \)[/tex].
6. Formulate the expression to prove it’s odd:
Notice that the terms [tex]\( 4n^2 \)[/tex] and [tex]\( 4n \)[/tex] are both even terms because they each have [tex]\( 4 \)[/tex] as a factor.
To express the entire expression in terms of even and odd parts:
[tex]\[
4n^2 + 4n + 1 = 2(2n^2 + 2n) + 1
\][/tex]
7. Conclude that the expression is odd:
The term [tex]\( 2(2n^2 + 2n) \)[/tex] is even because it’s a multiple of 2. When an even number is added to 1 (an odd number), the result is always odd.
Thus, the final expression is:
[tex]\[
2 \left(2n^2 + 2n\right) + 1
\][/tex]
This shows that the square of the sum of two consecutive integers [tex]\( (2n + 1)^2 = 4n^2 + 4n + 1 \)[/tex] is indeed odd. Therefore, we have proven that:
[tex]\[
(n + (n + 1))^2 = (2n + 1)^2 = 4n^2 + 4n + 1 = 2(2n^2 + 2n) + 1 \implies \text{odd}
\][/tex]
