Discrete Math Proof?

?

　
Let k = 0

((k(2k-1)(2k+1)) / 3) + (2k+1)^2
= ((0(-1)(1)) / 3) + (1)^2
= 0 + 1
= 1

(k+1(2k+1)(2k+2)) / 3
= (0+1(1)(2)) / 3
= 2/3

Therefore, it not possible to prove that ((k(2k-1)(2k+1)) / 3) + (2k+1)^2 is equal to (k+1(2k+1)(2k+2)) / 3

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

It does not equal what you propose.