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$$S_{n} = \sum_{i=0}^{n}(5i+3)$$

I received a homework problem that instructed me to use induction to prove that for all natural numbers n

$$S_{n} = \frac{n(5n+11)}{2}+3$$

First I proved that my base case of $S_{0}$ holds, because substituting $0$ for $n$ in both the top formula and the following formula makes both equal to $3$. The next step is to form my inductive hypothesis. My hypothesis is that

$$\sum_{i=0}^{n}(5i+3) = \frac{n(5n+11)}{2}+3$$ for all natural numbers $n$. Then I'm assuming that $$\sum_{i=0}^{k}(5i+3) = \frac{k(5k+11)}{2}+3$$ holds when $n$ = some arbitrary natural number $k$ (I've since been told not to do $n=k$ for some reason).

Next step is to prove that $S_{k+1}$ holds, because if it does, knowing that my base case holds will tell me that $S_{1}$ holds, telling me that $S_{2}$ holds, etc.

To prove this, I took the equation from my assumption and substituted $k+1$ for $k$. Evaluating the left hand side of $\frac{(k+1)(5(k+1)+11)}{2}+3$ eventually yielded $\frac{5k^2+21k+22}{2}$, and solving the right hand side of $\sum_{i=0}^{k+1}(5i+3)$ using Gauss's(?) sum and splitting the terms of the sum (I don't know what to call it) to come to the same result. Since both sides of the equation reduced to the same expression, I reasoned that this proves that my original assumption holds, therefore the statement at the top has been proven.

I've gone wrong somewhere above, since I was told that I proved the original assertion with a direct proof rather than by induction. Where did I go wrong? I thought that after making my assumption and learning the case that needs to hold to make such assumption true, all I need to do is see if both sides of the equation equal each other. Has doing a direct proof of the original statement caused me to make too many assumptions? Or have I done something else inappropriate?

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  • $\begingroup$ You were told... by whom? Your proof seems to line up with induction nicely. $\endgroup$ – abiessu Apr 4 at 0:16
  • $\begingroup$ @abiessu I was told this by my TA $\endgroup$ – user2709168 Apr 4 at 0:34
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Typically, you want to remember that, for proof by induction, you have to make use of the induction assumption. You assume some case greater than your base case holds, and then show it implies the succeeding step - that gives you the whole "$S_1 \implies S_2 \implies S_3 \implies ...$" chain.

So our assumption is

$$\sum_{i=0}^{k}(5i+3) = \frac{k(5k+11)}{2}+3$$

We seek to show

$$\sum_{i=0}^{k+1}(5i+3) = \frac{(k+1)(5(k+1)+11)}{2}+3 = \frac{(k+1)(5k+16)}{2}+3$$

Starting with the sum at the left, we can pull out the $(k+1)^{th}$ term:

$$\sum_{i=0}^{k+1}(5i+3) = 5(k+1) + 3 + \sum_{i=0}^{k}(5i+3) = 5k+8 + \sum_{i=0}^{k}(5i+3)$$

As it happens, this new summation is precisely what we assume holds. So we substitute the corresponding expression and do some algebra:

$$\begin{align} 5k+9 + \sum_{i=0}^{k}(5i+3) &= 5k+8 + \frac{k(5k+11)}{2}+3\\ &=\frac{10k+16 + 5k^2 + 11k}{2} + 3\\ &=\frac{5k^2+21k+16}{2} + 3\\ &= \frac{(k+1)(5k+16)}{2}+3 \end{align}$$

Thus, the case for $(k+1)$ holds, completing the induction step.

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  • $\begingroup$ I think I mixed up my expressions in the post, but my intention was to have what you had as your assumption(?) as my inductive hypothesis. Do I not use that hypothesis when proving that the k+1 substitution holds? $\endgroup$ – user2709168 Apr 4 at 0:48
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    $\begingroup$ You used something you refer to as "Gauss's (?) sum" in that, so, no, you did not make use of your induction hypothesis. At least in any obvious way because I have no idea what this sum you refer to is. $\endgroup$ – Eevee Trainer Apr 4 at 0:54
  • $\begingroup$ Are you saying my inductive hypothesis was Gauss's sum? Because that's not what I thought I was asserting Copy pasting a different comment of mine explaining what I meant: "I mentioned Gauss's sum because that's one of the things I used to evaluate the right side of my equation- through turning the sum of 5i into 5((k+1)(k+2))/2." $\endgroup$ – user2709168 Apr 4 at 0:56
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    $\begingroup$ I'm saying that you're not making use of the inductive hypothesis. You verified the inductive step by another method, which makes no use of the inductive hypothesis. You have to assume the inductive hypothesis holds when you verify the inductive step: that's the whole point of the "this implies that implies that" domino effect. Alongside the base case and the fact that one implies the next - and you have to have a step implying the next, and have to show that implication holds - that gives us the domino effect. Verifying the induction step independently does not show $S_k\implies S_{k+1}$. $\endgroup$ – Eevee Trainer Apr 4 at 1:04
  • $\begingroup$ What is my inductive hypothesis in this situation? I thought the hypothesis was that the statement holds when n=k, therefore by proving it holds for k+1 then it holds for all n $\endgroup$ – user2709168 Apr 4 at 1:08
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I think the place where you say you used the "Gauss sum" is where your instructor says you just gave a direct proof. It's hard to tell, because you didn't show us your proof, you just said "and then I did and then ...".

What's expected is that you write the result for a particular value of $k$ - the inductive hypothesis, then add the next term and do some algebra to show that you get the result for $k+1$.

As an aside, I really don't like a question that asks you to prove something by induction when there is an easier straightforward way - in this case, Gauss's method.

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  • $\begingroup$ What do you mean by a particular value of k? I mentioned Gauss's sum because that's one of the things I used to evaluate the right side of my equation- through turning the sum of 5i into 5((k+1)(k+2))/2. I thought the particular value was writing that the statement holds when the value of k is n (or the other way around? trying to figure out what supposedly went wrong is confusing me), and then I can prove I get the same result for k+1. $\endgroup$ – user2709168 Apr 4 at 0:53
  • $\begingroup$ I can't explain any better what I mean than what is in @EeveeTrainer 's answer and comments. $\endgroup$ – Ethan Bolker Apr 4 at 1:19

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