Induction for series of squares

[KOO]

Prove that for all nEN

1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = (4n^3 - n) / 3My Solution)

If n = 1, 4(1)^3 - 1 / 3 = 1 so base case holds.

Assume 1^2 + 3^2 + ... + (2k-1)^2 = (4k^3 - k) / 3

What next?

 

[Deveno]

Well, then, obviously:

$1^2 + 3^2 + 5^2 + \cdots + (2k - 1)^2 + (2k + 1)^2 = \dfrac{4k^3 - k}{3} + (2k + 1)^2$

$= \dfrac{4k^3 - k}{3} + \dfrac{12k^2 + 12k + 3}{3}$

$= \dfrac{4k^3 + 12k^2 + 12k + 4 - k - 1}{3}$

$= \dfrac{4(k+1)^3 - (k+1)}{3}$