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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.

Adam
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  • I recommend this answer as a great approach to induction. You have $S(k+1)=S(k)+(k+1)$ on the left. Take the formula for $S(k)$, add $k+1$ and get the formula you need for $S(k+1)$ – Ross Millikan May 12 '20 at 00:48

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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 $n=k+1$.

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

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

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

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

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

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

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