Solving $\sum_{i=3}^{n+1} i$ trying to solve $\sum_{i=3}^{n+1} i$
First I attempt to change the lower/upper bounds:
$\sum_{i=3}^{n+1} i = \sum_{i=1}^{n-1} i$ 
in order to use $\sum_{i=1}^{n} i = \frac{n(n+1)}2$
so, $\sum_{i=1}^{n-1} i = \frac{(n-1)(n-1+1)}2 = \frac{n^2-n}2$
This is just practice out of a textbook that doesn't have answers - but I tried to input the summation in wolframalpha and my result is not one of the answers there. Where have I messed up? 
Additionally, is modifying the lower/upper bound of a summation in order to use an equality like the one above an ok way to approach these problems?

That's wrong. Take $n=2$. Then you say $3=1$.
Use
$$
\sum_{i=3}^{n+1} i = \sum_{i=1}^{n+1} i-\sum_{i=1}^{2} i
$$
If you want change order of summation, then set $m=i-2$. Then $m$ goes from $1$ to $n-1$ but $i=m+2$ so
$$
\sum_{i=3}^{n+1} i=\sum_{m=1}^{n-1} (m+2)
$$
