Formula when index of sigma is negative We're currently learning series and sigma notation
We've been given the formulas for $\sum_{k=1}^{n}k$, $\sum_{k=1}^{n}k^2$, and $\sum_{k=1}^{n}k^3$ plus the properties on how to break them apart etc, place the constant c in front and multiply by the resulting sum.  
Now I've been given $\sum_{k=-1}^{5}k^2$  and I'm stuck because the formulas won't work for that.  If k=2 I understand that I can do $\sum_{k=1}^{5}k^2 - \sum_{k=1}^{1}k^2$ but that same concept won't work when k=-1
Is there a formula to solve it or is it just a matter of multiplying it all out 

I give you below the general formulae
$$\sum_{k=m}^{n}k=\frac{1}{2} (n-m+1) (n+m)$$
$$\sum_{k=m}^{n}k^2=\frac{1}{6} (n-m+1) \left(2 m^2+2 m n-m+2 n^2+n\right)$$
$$\sum_{k=m}^{n}k^3=\frac{1}{4} (n-m+1) (m+n) \left(m^2-m+n^2+n\right)$$
