Use mathematical induction to prove n ≥ 3 
I have proved for n =3, and assumed S(k) is true already. 
I have gotten all the way to the induction step of S(k+1) = 3+4+5+...+(k+1) = ((k+1-2)(k+1+3))/2
I am having trouble proving it past this step, and how to show that what I have added is equal on both sides. Please show me how I can do that last step of showing the truth.

Assume the formula holds for some $k\ge3$, that is:
$$S(k)= \frac{(k-2)(k+3)}{2}.$$
Now, $S(k+1) = S(k)+k+1$, so using the above formula we get:
$$S(k+1) = \frac{(k-2)(k+3)}{2} +k+1 = \frac{(k-2)(k+3)+2k+2}{2} = \frac{k^{2}+3k-4}{2} = \frac{(k-1)(k+4)}{2} = \frac{((k+1)-2)((k+1)+3)}{2}$$
which is your formula for $n=k+1$.
