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Could someone help me figure out where to begin with this proof by induction. Prove by induction that $$\sum_{i=1}^n (i+1) = \frac{(n)(n+3)}{2}.$$

Edit: I have proven that the base case n=1 is true as it will give you 2=2. Next I should complete the induction step but I'm weak in my understanding of the induction process. Any help would be greatly appreciated!

user489819
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    We could help you figure it out, but you need to give us somehting to work with, otherwise we're just giving you a proof, not helping you work it out. Have you checked any base cases? How are you doing with the induction step? – Arthur Nov 14 '17 at 14:20
  • Oh ok, I apologize I'm a newbie at this. Thank you for your response. I'm honestly not sure how to set up the proof in the first place. What do you think of setting the base case as i=1 resulting in the solution 2=2, so from there how do I prove that it's true for all n and not just 1? – user489819 Nov 14 '17 at 14:31
  • It's $n$ that you set equal to $1$, not $i$. But yes, in that case you would get $(1+1) = \frac{1(1+3)}2$, which turns out to be true, as you said ($2 = 2$). Cool. You have the base case down. Now for the induction step. How well do you know and / or understand the general procedure of an induction step proof? – Arthur Nov 14 '17 at 14:33
  • I understand it sort of, I believe the next step should be assume that since my base case was true that next I should (if my equation is represented by P) assume P(n) is true and solve for P(n+1)? – user489819 Nov 14 '17 at 14:51
  • And if I do that I get to P(n+1) = (n+1)(n+4)/2 but after that I'm not sure what to do. – user489819 Nov 14 '17 at 14:55

2 Answers2

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Because this will contain an answer, and is pretty long, I will continue here from the comments above.

You are right about the general idea about the induction step. To not mix up letters it's common to say "Assume $P(k)$ and prove $P(k+1)$" instead, but that's not really too important. Technically it is not correct to use $n$ in that phrase (and in the proof below), and you could use $k$ if you want, but at your level I would consider it pedantic to require you to use it.

At any rate, we're assuming that $$\sum_{i=1}^n(i+1)=\frac{n(n+3)}2$$(this is the so-called induction hypothesis) and we want to use it to show that $$\sum_{i=1}^{n+1}(i+1)=\frac{(n+1)(n+4)}2$$ So we start with the sum on left-hand side, and see what we can make of it. There is no point in assuming the induction hypothesis if we're not going to use it, so we try to massage our sum until we get something we can use. This time it's not so difficult: $$ \sum_{i=1}^{n+1}(i+1)=(n+1+1)+\sum_{i=1}^{n}(i+1) $$ Now we can use the induction hypothesis (note that the sum on the right-hand side is identical to the sum in the induction hypothesis), and substitute the sum on the right side with the appropriate fraction: $$ (n+1+1)+\sum_{i=1}^{n}(i+1)=n+2+\frac{n(n+3)}2 $$ All that is left is to see whether this right-hand side is actually equal to $\frac{(n+1)(n+4)}2$. If it is, then we have proven our $P(n+1)$, and by induction, the formula is true for all natural numbers.

Arthur
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  • Arthur thank you so much for the detailed explanation! That really helped me to understand it. I really appreciate it! – user489819 Nov 14 '17 at 15:13
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Consider the statement $$P(n): \sum_{i=1}^n(i+1)=\frac{n(n+3)}{2}$$

Then for $n=1$, $1+1=\frac{1(1+3)}{2}$ i.e $2=2$. Hence $P(1)$ is true.

For $n=2$;

LHS$=(1+1)+(2+1)=5=\frac{2(2+3)}{2}=$RHS

Therefore $P(2)$ is also true.

Induction hypothesis: Suppose $P(n)$ is true for $n=k.$ i.e $$P(k): \sum_{i=1}^k(i+1)=\frac{k(k+3)}{2}$$ is true.

Induction step:( to show $P(n)$ is true for $n=k+1$).

$$P(k+1)=\sum_{i=1}^{k+1}(i+1)=\sum_{i=1}^{k}(i+1)+(k+1)=P(k)+(k+1)+1$$ $$=\frac{k(k+3)}{2}+(k+1)+1$$ $$=\frac{k(k+3)+2(k+1)+2}{2}$$ $$=\frac{(k+1)(k+1+3)}{2}$$

Hence $P(k+1)$ is true.

Dastan
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