The value of $$1^2 + 3^2 + 5^2 + \ldots + 25^2$$ is :

= $$\sum_{n=1}^{13} (2n-1)^2$$

$$(2n-1)^2 = 4n^2 - 4n + 1$$

$$\sum_{n=1}^{13} (4n^2 - 4n + 1) = 4\sum_{n=1}^{13} n^2 - 4\sum_{n=1}^{13} n + \sum_{n=1}^{13} 1$$

$$\sum n^2 = \frac{k(k+1)(2k+1)}{6} = \frac{13 \times 14 \times 27}{6} = 819$$

$$\sum n = \frac{k(k+1)}{2} = \frac{13 \times 14}{2} = 91$$

$$\sum 1 = k = 13$$

$$\text{Total} = 4(819) - 4(91) + 13$$

$$\text{Total} = 3276 - 364 + 13$$

$$\text{Total} = 2912 + 13 = \mathbf{2925}$$