Question

(d) 2n + 1 is an odd number. Show that the product of 2 consecutive odd numbers is always an odd number

Answer 1

$n^{2}$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$n^{2}$+6n + 2n +3

= 4$n^{2}$+8n +3

What we can do now is separate the +3 into +1 and +2

4$n^{2}$+8n +3

4$n^{2}$+8n +2 +1

Now we can factor out a 2 from the equation

2(2$n^{2}$+4n +1) +1

Now you can see that we have the equation:

2 times some number + 1

No matter what (2$n^{2}$+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.