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.

$$S(n)+(n+1)=$$
$$3+4+5+...+n +(n+1)=$$
$$\frac{(n-2)(n+3)}{2}+ (n+1)=$$
$$\frac{(n-2)(n+3)+2(n+1)}{2}=$$
$$\frac{n^2+n-6+2n+2}{2}=$$
$$\frac{n^2+4n-n-4}{2}=$$
$$\frac{n(n+4)-(n+4)}{2}=$$
$$\frac{(n-1)(n+4)}{2}=$$
$$\frac{((n+1)-2)((n+1)+3)}{2}=$$
$$S(n+1)$$
