Is ∑ with negative value solvable? Is it possible to have a negative value in sigma?
e.g.
$y = \Sigma_{k=0}^{k=-2} k \times 10$   
Will this give the result $(0 \times 10) + (-1 \times 10) + (-2 \times 10) = -30 $?
Or will it be $\infty$ because $k$ will be increased with $1$ until it equals $-2$ (which is never).
Or something else?

The concept of Sum has three basic definitions.

*

*Sum over  (part of) a sequence
Given a unilateral sequence
$$
x_{\,0} ,\,x_{\,1} ,\, \cdots ,\;x_{\,n} 
$$
we define a sum over a portion of it as
$$
\sum\limits_{k = \,a}^b {x_{\,k} } 
$$
where it is understood that either $a$ and $b$ are integers and that $a \le b$.
Under this acception your sum does not have meaning.
Another way of writing the sum is by imposing restriction to the index
$$
\sum\limits_{a\, \le \,k\, \le \,b} {x_{\,k} } 
$$
and if the condition is violated the sum is null.
But the sequence could be bi-lateral
$$
 \cdots ,\;x_{\, - n} ,\; \cdots x_{\, - 1} ,x_{\,0} ,\,x_{\,1} ,\, \cdots ,\;x_{\,n} ,\; \cdots 
$$
In this case you may want to write
$$
\sum\limits_{k = \,0}^b {x_{\,k} } 
$$
leaving $b$ free to address any index in the sequence, and thus understanding
$$
\sum\limits_{k = \,0}^b {x_{\,k} }  = \sum\limits_{k = \,b}^0 {x_{\,k} } 
$$
but you shall clearly state this acception.


*Sum over a set
$$
\sum\limits_{x\, \in \,A} x 
$$
and the meaning is clear.


*Indefinite sum (Antidelta)
Finally there is the concept of Indefinite Sum.
For a function $F(z)$, over the complex field in general, we define the (forward) Finite Difference as
$$
\Delta \,F(z) = F(z + 1) - F(z)
$$
If we have that
$$
\Delta  \,F(z) = F(z + 1) - F(z) = f(z)
$$
then we write
$$
F(z) = \Delta ^{\, - \,1} \,F(z) = \sum\nolimits_{\;z\;} {f(z)} + c
$$
and in particular we have
$$
F(b) - F(a) = \sum\nolimits_{\;z = \,a\,}^b {f(z)}  =  - \sum\nolimits_{\;z = \,b\,}^a {f(z)} \quad \left| {\,a,b \in \mathbb C} \right.
$$
For example
$$
\eqalign{
  & F(z) = \left( \matrix{  z \cr  2 \cr}  \right)\quad  \Leftrightarrow \quad
 \left( \matrix{  z + 1 \cr   2 \cr}  \right) - \left( \matrix{  z \cr   2 \cr}  \right)
 = \left( \matrix{  z \cr   1 \cr}  \right) = z\quad  \Leftrightarrow   \cr 
  &  \Leftrightarrow \quad \sum\nolimits_{\;x = \,a\,}^{\;b} z
  = \left( \matrix{  b \cr   2 \cr}  \right) - \left( \matrix{  a \cr   2 \cr}  \right)
 \quad \left| {\;a,b \in \mathbb C} \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \sum\nolimits_{\;k = \,0\,}^{\;n} k \quad \left| {\;0 \le n \in \mathbb Z} \right.\quad
  = \sum\limits_{0\, \le \,k\, \le \,n - 1} k  = \left( \matrix{  n \cr   2 \cr}  \right)
 = {{n\left( {n - 1} \right)} \over 2} \cr} 
$$
