someone please help with this question ?:)?

Question

1,3,5,7

The first odd number can be expressed as
1=1²-0²

The second odd number can be expressed as 3=2²-1²

The third odd number can be expressed as
5=3²-2²

The fourth odd number can expressed as
7=4²-3²

Write down a formula for the Nth odd number and simplify this expression & prove that the product of two consecutive odd numbers is always odd.

could you show me how you did it aswell please? very confused!
xo

Answer 1

Formula is (n+1)^2-n^2: (n+1)^2=n^2+2n+1, so (n+1)^2-n^2=(n^2+2n+1)-n^2=2n+1 is always odd.

The product of any two odds, not just two consecutive odds, is odd. (2n+1)(2k+1)=4nk+2k+2n+1=2(2nk+k+n)+1 is odd.

Answer 2

nth odd number = n^2 - (n-1)^2
= n^2 - n^2 + 2n - 1
= 2n - 1

verification; when 'n' = 15 then 2(15) - 1 = 29 (odd).

Product:-

(n^2 - (n-1)^2)( (n+1)^2 - n^2)
(n^2 - n^2 +2n -1)(n^2 + 2n + 1 - n^2)
If 2n -1 is odd then the next consecutive odd number is 2n + 1
Multiplying together
(2n-1)(2n+1)
4n^2 - 2n + 2n - 1
4n^2 - 1 This answer is always odd, because if 'n' is odd/even then 'n^2' will be odd/even.However, '4' is even and 'n^2' is odd/even and an even multiplied to an odd/even always gives an even answer. Therefore 4n^2 is always even. Hence subtracting '1' from an even number will result in an 'odd' number.

Verification.

If 'n' = 5 then 4(5)^2 - 1 = 99
If 'n' = 6 then 4(6)^2 - 1 =143
Try it for yourself.
NB Vertifications are NOT proofs. Using algebra and 'n' is the proof.