- Thread starter
- #1

- Thread starter KOO
- Start date

- Thread starter
- #1

- Feb 15, 2012

- 1,967

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

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

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

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