Mathmatical Induction

1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 = [n(2n - 1)(2n + 1)]/3

First....let us show it is true for, n = 1......that is.....

1^2   =  [ (1) ( 1) ( 3) ] / 3   =  [ 3/3]   = 1     

1^2  = 1      and it's true......

Now, assume that it's true for  n = k  ........that is......

1^2 + 3^2 + 5^2 + ... + (2k - 1)^2 = [k(2k - 1)(2k + 1)]/3     (1)

So, prove that it's true for n = k + 1  .......that is

1^2 + 3^2 + 5^2 + ...+  (2k-1)^2 + (2(k+1) -1)^2 = [(k+ 1)(2k+ 1)(2k + 3)]/3 ...or...put more simply...

1^2 + 3^2 + 5^2 + ... + (2k - 1)^2 + (2k + 1)^2 = [ (2k + 1) (2k + 3) (k + 1) ] / 3

....so we have.....

[Add (2(k+ 1) -1)^2  to  both sides of (1)  ]

1^2 + 3^2 + 5^2 + ...  + (2k - 1)^2  + (2 (k + 1) -1)^2 = [k(2k - 1)(2k + 1)]/3  + (2(k + 1) - 1)^2

Simplify

1^2 + 3^2 + 5^2 + ... + (2k - 1)^2 + (2k + 1)^2 = [k(2k - 1)(2k + 1)]/3  + (2k + 1)^2

Get a common denominator on the right side

1^2 + 3^2 + 5^2 + ... + (2k - 1)^2 + (2k + 1)^2 = [ (k) (2k-1) (2k + 1) + 3(2k + 1)^2 ] /3

Factor out (2k + 1)  in the numerator on the right

1^2 + 3^2 + 5^2 + ... + (2k - 1)^2 + (2k + 1)^2 = [ (2k + 1) [ k (2k - 1) + 3(2k + 1)] ]/ 3

Simplify that numerator

1^2 + 3^2 + 5^2 + ... + (2k - 1)^2 + (2k + 1)^2 = [ (2k + 1) ( 2k^2 - k + 6k + 3) ] / 3

1^2 + 3^2 + 5^2 + ... + (2k - 1)^2 + (2k + 1)^2 = [ (2k + 1) (2k^2 + 5k + 3) ] / 3

Factor (2k^2 + 5k + 3 )  as (2k + 3) (k + 1)

1^2 + 3^2 + 5^2 + ... + (2k - 1)^2 + (2k + 1)^2 = [ (2k + 1) (2k + 3) (k + 1) ] / 3

Which is exactly what we wanted to prove......