\sum_{k=1}^{m}(2k-1)^2=\sum_{k=1}^{2m}k^2-\sum_{k=1}^{m}(2k)^2=\frac{m(2m+1)(4m+1)}{3}-\frac{2m(m+1)(2m+1)}{3}=\frac{m(2m-1)(2m+1)}{3}