Find the sum of the series:
(a) (1^2)+(3^2)+(5^2)+...+(2n-1)^2
(b) (1^2)-(2^2)+(3^2)-(4^2)+...+((-1)^(n+1))(n^2)
a)
$\displaystyle \displaystyle 1^{2} + 3^{2} + 5^{2} + ...+ (2n-1)^{2} = $
$\displaystyle \displaystyle = 1^{2} + 2^{2} + 3^{2} + ...+ (2n-1)^{2} - 2^{2}\ \{1^{2} + 2^{2} + 3^{2} + ...+ (n-1)^{2}\} =$
$\displaystyle \displaystyle = \frac{2n\ (2n-1)\ (4n-1) - 4n\ (n-1)\ (2n-1)}{6} = $
$\displaystyle \displaystyle = \frac{2n\ (2n-1)\ (2n+1)}{6}$
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$