Is there a quick way to prove the following? The method I used seems bit too long and lengthy, which I'd rather not post here.
$$\sum_{t=1}^n t^2=1^2+2^2+3^2+\dots+t^2=\frac{n(n+1)(2n+1)}6$$
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When you add $(n+1)^2$, bring to the common denominator $6$, "take out" a common factor $n+1$, and the remaining algebra is pretty quick. – André Nicolas Jul 27 '16 at 23:45
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See http://math.stackexchange.com/questions/48080/prove-that-sum-limits-k-1nk2-fracnn12n16 and http://math.stackexchange.com/questions/435412/induction-proof-sum-k-1nk2-fracnn12n16 – Martin Sleziak Jul 28 '16 at 05:11
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Check that it's true for $1 \le n \le 4$. Then observe by taking differences that the left-hand side of the equation is a polynomial and that that polynomial has degree at most $3$. If two polynomials of degree at most $3$ agree on $4$ values, then they are equal. See https://www.math.upenn.edu/~wilf/AeqB.html for much, much, more along these lines.
Rob Arthan
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2I was about to say this reminds me a great deal of something my professor would do, and then I see you've linked his book! – Sidharth Ghoshal Jul 27 '16 at 23:52
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1@frogeyedpeas: The professor in question (Doron Zeilberger, I assume) is a very cute cookie! I envy you for having him as a teacher. – Rob Arthan Jul 28 '16 at 00:00
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I assume the equation is meant to be $\sum_{t=1}^{n}t^2 = 1^2 + 2^2 + \dots + \color{red}{n}^2$ instead of $\sum_{t=1}^{n}t^2 = 1^2 + 2^2 + \dots + \color{red}{t}^2$.
The equation holds for $n=1$ because $$\sum_{t=1}^{1} = 1^2 = 1$$ and $$\frac{1(1+1)(2\times 1+1)}{6} = 1$$ so $$\sum_{t=1}^{1} = \frac{1(1+1)(2\times 1+1)}{6}$$
Now if the equation holds for $n = k$ where $k \in \mathbb{N}_{\ge 1}$, then $$\sum_{t=1}^{k} = 1^2 + 2^2 + \dots + k^2 = \frac{k(k+1)(2k+1)}{6} = \frac{2k^3 + 3k^2 + k}{6}$$ and hence $$\sum_{t=1}^{k+1} = 1^2 + 2^2 + \dots + k^2 + (k+1)^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2 = \frac{2k^3 + 3k^2 + k}{6} + (k^2 + 2k +1) = \frac{2k^3 + 3k^2 + k}{6} + \frac{6k^2 + 12k+6}{6} = \frac{2k^3+9k^2+13k+6}{6}$$
Also $$\frac{[k+1]([k+1]+1)(2[k+1]+1)}{6} = \frac{(k+1)(k+2)(2k+3)}{6} = \frac{2k^3+9k^2+13k+6}{6}$$ so $$\sum_{t=1}^{k+1}t^2 = \frac{[k+1]([k+1]+1)(2[k+1]+1)}{6}$$ and thus the equation holds for $n=k+1$.
Therefore by induction $$\sum_{t=1}^{n}t^2 = 1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}$$
As a straightforward induction proof, it cannot get any shorter.
Hope it helps. :)
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No be honest, I don't know how to phrase such a proof. That's why I put a disclaimer at the end. I've never really learned how to do mathematical proofing, but it I guessed I should start somewhere. As I understand it, you proof that the formula is valid for a certain natural number, let's say $1$. Then you proof that if the formula is vald for $n = k$, it is also valid for $n = k+1$. And since you have proven that it is valid for $n = 1$ and that it is valid for every natural number after a valid one, you have proven that it is valid for $n = 2$. (Part 1/2) – Kevin Jul 31 '16 at 02:13
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And using the same reasoning, since it is valid for $n = 2$, it is also valid for $n = 3$ and so forth. Basically,you have proven that it is valid for every number in the set ${ n | n \in \mathbb{Z, n \geq 1 }}$. Is this correct? I would be really grateful if you could tell what I'm doing wrong and tell me how I can correct them. (Part 2/2) – Kevin Jul 31 '16 at 02:21
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Excellent. You do understand induction. I've edited your answer to make the reasoning clearer. See if you like it. There is only one minor point that you should be aware of. Your justification of induction is nearly complete except that you need to know that although you can prove some statement "$P(n)$" for every natural number $n$ of the form "$1+1+\cdots+1$", it does not automatically mean you have proven a statement of the form "$\forall n \in \mathbb{N}\ ( P(n) )$". In fact, because of a deep result of logic that Godel discovered, there are cases where this fails. [cont] – user21820 Jul 31 '16 at 02:30
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[cont] This also explains why we need the induction axiom, because we cannot otherwise infer the universally quantified statement. Notice that without induction the proof of "$P(n)$" is different for each $n$ and gets longer as $n$ increases. But with induction we only need to prove "$P(0)$" and "$\forall n\in\mathbb{N}\ ( P(n) \to P(n+1) )$" and then we conclude by the induction axiom that "$\forall n\in\mathbb{N}\ ( P(n) )$". We can 'see from outside the proof world' that this axiom is valid when every natural number is of the form "$1+1+\cdots+1$", and that's why we're fine with adding it. – user21820 Jul 31 '16 at 02:33
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For reference, a more detailed explanation of what I just described in my comments is at http://matheducators.stackexchange.com/a/10033. Feel free to ask further if you've any questions! =) – user21820 Jul 31 '16 at 02:36
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Thanks for the explanation. This is how I understand the induction axiom: Say you have a set $S \subset \mathbb{N}$ which contains $1$. Also, for every element with value $k$ in this set, the element with value $k + 1$ is also in this set. Now, by the induction axiom $S = {n | n \in \mathbb{N}, n \neq 0 }$. Is my understanding correct? And if you give a proof by induction, can you use this induction axiom as a given? Also, what is the deep result in logic that Godel discovered? Thanks in advance. :) – Kevin Jul 31 '16 at 13:57
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What you say is true, and you can prove it by induction, but induction itself is the rule that from $\def\nn{\mathbb{N}}$"$P(0)$" and "$\forall n \in \nn\ ( P(n) \to P(n+1) )$" we can derive "$\forall n \in \nn\ ( P(n) )$". Usually $0 \in \nn$ and induction begins from $0$. To really grasp what is going on you have to learn logic (try the first two references I give in http://math.stackexchange.com/a/1684208) and natural deduction (see my profile). Then you'll understand my post about induction properly. Eventually you'll come to the deep result, Godel's incompleteness theorem. – user21820 Jul 31 '16 at 14:52