Use mathematical induction to prove n ≥ 3

Question

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.

Answer 1

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 $n=k+1$.

Answer 2

$$S(n)+(n+1)=$$

$$3+4+5+...+n +(n+1)=$$

$$\frac{(n-2)(n+3)}{2}+ (n+1)=$$

$$\frac{(n-2)(n+3)+2(n+1)}{2}=$$

$$\frac{n^2+n-6+2n+2}{2}=$$

$$\frac{n^2+4n-n-4}{2}=$$

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