2

So, I have to prove this by mathematical induction and I have never done it!

$$\sum_{i=1}^n i^2 = n(n+1)(2n+1)/6$$

However, I have learnt better the theory behind this way to prove statements and studied a couple of easy examples yesterday. I am trying this now but it may be I make some mistakes - perhaps, at the beginning - because I don't get an identity in the final passage. Is this the goal, right? Namely, two have 2 equal lines in the equation (?)

This is my attempt:

Let's prove the base case before with 2 (I choose 2 instead than 1 because with 1 the formula looks more trivial, in my opinion. I guess the choice of the number is arbitrary). Thus:

n=2 , i = sum of natural numbers, according to the rule, up to n, in this case, up to 2. I get:

$$1^2 + 2^2 = 2(2+1) (2*2+1) / 6$$ $$5 = 5$$ (I jumped a step in the computation here to be shorter)

So, the formula is true for 2.

Now, I have to introduce a general number k such that k is a natural number and, for underlying the progression of N, I write: k+1.

However, at this point I have my first doubt: how to write i^2?. Should I leave it like this and replacing only n by k+1?. I am trying to find a way for writig also i in terms of k+1 but I can't find it. So, for now I leave it like this and I work only on the right side of the formula. As I have understood so far, I have to replace n by k+1 simply. So, I write only the right part of the formula in this way:

$$(k+1)*[(k+1)+1]*[2(k+1)+1] / 6$$

I compute the terms:

$$(k+1)*(k+2)*(2k+3) / 6$$

I expand:

$$k^2+2k+k+2*(2k+3) / 6$$

I sum like terms:

$$(k^2+3k+2)*(2k+3) / 6$$

Expand again:

$$2k^3+3k^2+6k^2+9k+4k+6 / 6$$

Sum like terms again:

$$(2k^3+9k^2+13k+6) / 6$$

Already in this passage I don't like this 13, prime number which doesn't allow me to simplify nicely, and which let me think that I made a mistake. I could simplify in this way:

$$k^3/3 + 3k^2/2 + 13k/6 + 1$$ but I don't end up with any similarity with the left side of the formula.

Could you give me some help, please? ...I am already tired...

UPDATE:

I hope you don't put this post as a duplicate for the fact is going to be proved the same statement, as I am trying also to learn how to write a proof with my own style. So, answers to this question may help me to achieve the goal, at least for this question. However, I read and I will also similar questions already posted in order to learn more about proof writing.

UPDATE: Can I write the proof in the following way? After proved for the base case, I write:

Let's assume that k=n and there exists a general number k+1 which appartains at N (I would like to condensate what you refer as steps 2 and 3) so that $$\sum_{i=1}^n i^2 = \sum_{i=1}^{i=k+1} i^2 = k(k+1)(2k+1)/6 + (k+1)^2$$ - I don't know how to put that 1 upon sigma

and then I simply proceed with the algebraic passages (?)

BLAZE
  • 8,458
Always learning
  • 633
  • you need to use an assumption: so step 2 is generally, assume that $n=k$ and that the sum equals $\frac{k(k+1)(2k+1)}{6}$. Then for $n+1$ you use the sum formula $\sum_{i=0}^{k}i^2+(k+1)^2$ and now show that it equals $\frac{(k+1)(k+2)(2k+3)}{6}$ – Eleven-Eleven Oct 26 '15 at 09:53
  • Thank you! So, working on the left side of the formula, I would have: $$i^2 + k^2+2k+1$$ and I should have a similar line on the right side of the formula? – Always learning Oct 26 '15 at 09:58
  • no, you have $$\frac{k(k+1)(2k+1)}{6}+k^2+2k+1$$and you want to show that equals the right hand side i mentioned above – Eleven-Eleven Oct 26 '15 at 10:05
  • no problem. Just remember that, under the $n=k$ assumptions you want the sum up to $k+1$, which is $$\sum_{i=1}^{k+1}{i^2}=\sum_{i=1}^k{i^2}+(k+1)^2$$. Now just show that the expression is equivalent to when you plug $(k+1)$ into the formula you are trying to prove. – Eleven-Eleven Oct 26 '15 at 10:13
  • 5
    You may want to look at the following: Prove that $1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ for $n \in \mathbb{N}$. – Daniel W. Farlow Oct 26 '15 at 10:13
  • @DanielW.Farlow, thank you for pointing me out this post. I have already read it and I will examine with more attention later (just because at the moment I feel tired). – Always learning Oct 26 '15 at 11:18
  • You should do the base case for $n=1$, and not for $n=2$, as you need to prove the statement also for $n=1$, not just $n=2$ and up. Also, as you stated, $n=1$ is a really easy case to deal with, and you always want things to be as simple as possible when dealing with mathematics. It's already hard enough as it is when you don't make an effort to complicate matters. – Scounged Oct 26 '15 at 11:23
  • I get your point, thank you! However, I just thought like this because, from a psychological standpoint, checking if the hypothesis holds for another number, can help in believing in the statement and doing the effort of proving it. Of course, I know that this statement must be true...because it is given in an exercise...but, at any rate, this is why I tried with 2 instead than 1. – Always learning Oct 26 '15 at 11:29
  • @Eleven-Eleven, would you like to show me all passages, perhaps writing an answer to the post? – Always learning Oct 26 '15 at 12:56
  • @DanielW.Farlow, you are perhaps the one with more experience, could you tell me how factorization is performed from 2k^3+9k^2+13k+6 to (k^2+3k+2)(2k+3), please? – Always learning Oct 26 '15 at 14:13
  • @Alwayslearning Where in my answer do you see me try to factor that? I avoided that route for that very reason--you run into an unnecessary complication. Is there anything about my answer that does not make sense to you? – Daniel W. Farlow Oct 27 '15 at 15:22
  • @Alwayslearning the user Eleven-Eleven has given a detailed answer and he has also answered all your questions ,if you haven't more objections you should accept his answer. – Nameless Oct 28 '15 at 08:21
  • @Alwayslearning I wanted to post an answer of my own here, but I cannot as the thread has already been marked as a duplicate. So I have posted an answer here instead as I think I can explain it to you in the same way we discussed before. But for this thread you have an excellent answer and might as well accept it (and since no one else can answer here now). – BLAZE Oct 28 '15 at 08:22

1 Answers1

1

Induction is a 3 step process. The first step will always be the base case. So, assuming induction on the natural numbers or some subset of the natural numbers, there will always be a least element that must satisfies the expression requiring proof. For your case, you need to show that $$\sum_{i=1}^1{i^2}=1^2=1$$ and $$\frac{1(1+1)(2\cdot 1+1)}{6}=\frac{1\cdot 2\cdot 3}{6}=\frac6{6}=1$$ Thus, for $n=1$, this formula for the sum of natural number squares holds. Now for induction, we ASSUME that the case where $n=k$ is true. Notice here that $k$ is assumed to be ANY natural number bigger than 1 (since we wanted the lowest such number to be true as our base case). Under this assumption then, $$\sum_{i=1}^ki^2=1^2+2^2+...+k^2=\frac{k(k+1)(2k+1)}{6}$$ Finally, the last step is to show that this particular formula holds when we replace $k$ with $k+1$. This method is sometimes called the domino proof because if you consider assuming true a statement for some $n$ and it is then true for the next number, it is then true for all natural numbers, since our choice of $n$ is arbitrary. To show the proof, start with $$\sum_{i=1}^{k+1}=1^2+2^2+...+k^2+(k+1)^2$$ Now we already have a formula for the sum of the first $k$ so we can replace that section with the formula: in other words $$\sum_{i=1}^{k+1}=1^2+2^2+...+k^2+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$$ Now we finish. We need to manipulate our right hand side expression to display the fact that we replaced $k$ with $k+1$, so our goal is to show that $$\frac{k(k+1)(2k+1)}{6}+(k+1)^2=\frac{[(k+1)[(k+1)+1)][2(k+1)+1]}{6}$$ It is easier not to simply multiply out. Leave the LHS in factored form $$\frac{k(k+1)(2k+1)}{6}+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+\frac{6(k+1)^2}{6}$$ $$=\frac{(k+1)[k(2k+1)+6(k+1)]}{6}$$ $$=\frac{(k+1)(2k^2+7k+6)}{6}$$ $$=\frac{(k+1)(2k^2+4k+3k+6)}{6}$$ $$=\frac{(k+1)[2k(k+2)+3(k+2)]}{6}$$ $$=\frac{(k+1)(k+2)(2k+3)}{6}$$ $$=\frac{(k+1)[(k+1)+1][2(k+1)+1]}{6}$$ This completes the proof.

Eleven-Eleven
  • 8,806
  • I understand it 100% for the first time! Thank you! Very clear, tidy and essential! – Always learning Oct 26 '15 at 21:25
  • Just for learning more, for what the acronym LHS stands for? – Always learning Oct 26 '15 at 22:35
  • I am writing it in order to understand more and I notice that in the second line, after you wrote: "Leave the LHS in factored form", you missed the square at the parenthesis (k+1) – Always learning Oct 26 '15 at 22:46
  • So, after that, is wrong what you wrote - because you ignored the square at the parenthesis (k+1). – Always learning Oct 26 '15 at 22:55
  • I wait for your edits before choosing your answer. I find interesting your advice not to expand the parenthesis as then we have to factor again and, although I have learnt how to do it, it's not easy to do (therefore, it is useless to expand). – Always learning Oct 26 '15 at 23:03
  • It is not wrong. Both sides have (k+1) as a common factor...it has been factored out and condensed. You now have a much easier factorization as well. If you can't factor easily,mathematics will become very difficult very quick. You need to be able to handle these easily. – Eleven-Eleven Oct 26 '15 at 23:23
  • Also note by factoring out the k+1, you avoid factoring a cubic and only have a quadratic which is is quite easy. The proof is correct as I have done this many times. – Eleven-Eleven Oct 26 '15 at 23:28
  • Thank you for your answer. I believe you but could you show me and explain me how you have factored out and condensed, please? There should be an intermediate step for doing it... – Always learning Oct 27 '15 at 08:29
  • Note that $$k(k+1)(2k+1)+6(k+1)^2=(k+1)[k(2k+1)]+(k+1)[6(k+1)]$$ and thus both LHS and RHS have a common factor of $(k+1)$. So by the distributive property we get $$(k+1)[k(2k+1)+6(k+1)]=(k+1)[2k^2+7k+6]$$ which is easily factored by any elementary methods taught in algebra 1. – Eleven-Eleven Oct 28 '15 at 02:20