Formula when index of sigma is negative

Question

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

Answer 1

Hint: We have \begin{align*} \color{blue}{\sum_{k=-1}^5k^2}&=(-1)^2+0^2+1^2+\cdots+5^2\\ &=\sum_{k=-1}^0k^2+\sum_{k=1}^5k^2\\ &\,\,\color{blue}{=1+\sum_{k=1}^5k^2}\\ \end{align*}

Answer 2

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