\sum_{l=1,3,5,...,n}(2l+1)=\sum^{\frac{n-1}{2}}_{p=0}(4p+3)=4\sum^{\frac{n-1}{2}}_{p=0}p+3\sum^{\frac{n-1}{2}}_{p=0}1=\frac{(n+1)(n+2)}{2}