[tex]n^{2}[/tex]First, understand that 2n+1 means that 2 times any number plus 1 will be an odd number. You can try this out. If you plug in any number, multiply it by 2 and add 1 it will be odd.
Second, consecutive odd integers are always two numbers apart.
3,5,7,9, etc...
3+2 = 5
5+2 = 7
7+2 = 9 etc...
So two consecutive odd integers can be written as:
2n+1 and (2n+1) +2
we can simplify (2n+1) +2 into 2n +3
the product of two consecutive odd integers would be
(2n+1) x (2n+3)
if we distribute this we get:
4[tex]n^{2}[/tex]+6n + 2n +3
= 4[tex]n^{2}[/tex]+8n +3
What we can do now is separate the +3 into +1 and +2
4[tex]n^{2}[/tex]+8n +3
4[tex]n^{2}[/tex]+8n +2 +1
Now we can factor out a 2 from the equation
2(2[tex]n^{2}[/tex]+4n +1) +1
Now you can see that we have the equation:
2 times some number + 1
No matter what (2[tex]n^{2}[/tex]+4n +1) is, if we multiply it by 2 and add 1, it will be odd.
So we have proven that the product 2 consecutive odd integers is odd.
