Computing $\sum_{i=2}^{n-1} 1$ and $\sum_{i=2}^{n-1} i$. Could someone explain how I calculate these summations? I'm using upper - lower + 1. 
a) $\displaystyle\sum_{i=2}^{n-1} 1$
b) $\displaystyle\sum_{i=2}^{n-1} i$
So for (a) I have: 
$$
(n-1) -2 +1)1 = -1n - 1.
$$
For (b) I have:
$$
((n-1) -2 +1)2 = -2n -1.
$$
Are these calculations correct?

For a). You go from $2$ to $n-1$ and add each time the value $1$. Thus you are summing up $$(n-1)-1$$ terms each equal to 1. Therefore $$\sum_{i=2}^{n-1}1=(n-2)\cdot1=n-2$$ (plug in some values of $n$. f.e. $n=3, n=4$ etc. to verify that this is correct. 
For b). On way to do it is to use the well known formula of the sum of the first $n$ consequtive integers that is equal to $n(n+1)/2$. Now, you can write the given sum as $$\sum_{i=2}^{n-1}i=\left(\sum_{i=2}^{n-1}i\right)\pm1\pm n=\sum_{i=1}^{n}i-(1+n)$$ and use the above formula to obtain that $$\sum_{i=2}^{n-1}=\sum_{i=1}^{n}i-(1+n)=\frac{n(n+1)}{2}-(n+1)=(n+1)\left(\frac{n}{2}-1\right)$$ 
