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Using Proof By Induction I am trying to prove the following:

$n^2 = \sum_{i=1} ^{n} (2i-1) $ for all $n\geq 1$

Here is my solutions so Far:

Base Case: $n=1, LHS: 2(1)-1 = 1, RHS = 1^2 = 1, True$

Induction Hypothesis:

Assume true for $n=k$

$k^2 =\sum_{i=1} ^{k} (2i-1) $ for some $k\geq 1$

Induction Step: Should be True for $n=k+1$

$(k+1)^2 =\sum_{i=1} ^{k+1} (2i-1) $ for some $k\geq 1$

However here is where I get stuck

RandomMath
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  • Are you only allowed an inductive proof? Perturbation method would be much better – Alex Apr 26 '15 at 14:26
  • @Alex Yes only Inductive – RandomMath Apr 26 '15 at 14:27
  • On your hypothesis step you said it was true for all $k \geq 1$. If that was true then you are done. You should instead say it is true for SOME $k \ge 1$. – Ebearr Apr 26 '15 at 14:27
  • @Ebearr - Changed – RandomMath Apr 26 '15 at 14:29
  • In the original question, the summation should be taken from $i=1$ instead of from $i=0$. Putting $n=1$ clearly shows this. – Hypergeometricx Apr 26 '15 at 14:35
  • But this question is better, and is asking for help completing a given attempted proof, which is not the case for the other question. – Jonas Meyer Apr 26 '15 at 16:24
  • @JonasMeyer I assume you are referring to the duplicate question. While it is true that this question does show some previous work (and thus it is better), the accepted answer is close to the best possible. – A.P. Apr 28 '15 at 16:23
  • @A.P., Yes, I think my comment made more sense in context when the preceding one read "possible duplicate of..." (now autodeleted). My reasons for favoring not having this closed as duplicate of the other include what you mention, but also include my opinion that it is not a duplicate due to the fact that it is asking for feedback/help with a particular attempt of the problem, rather than just asking for solutions to the problem. If too few others agree, oh well. In either case, it's good that there's a link to the other. – Jonas Meyer Apr 28 '15 at 16:31
  • @JonasMeyer This old discussion on Meta agrees with your point when it comes to proof verification (and I do, too). I still think that questions asking for proof completion should be closed when the completed proof (not just any proof) has already been posted elsewhere on MSE, because any answer would invariably provide a complete proof. – A.P. Apr 28 '15 at 16:51
  • @A.P.: Thank you, that distinction makes sense, even though there is some overlap. – Jonas Meyer Apr 28 '15 at 16:51

1 Answers1

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We want to show that, given the inductive hypothesis, $$(k+1)^2 =\sum_{i=1} ^{k+1} (2i-1)$$

LHS: $k^2 + 2k+1$.

The RHS

$\begin{align} \sum_{i=1} ^{k+1} (2i-1)& \overset{IH}{=} k^2 + 2(k+1) - 1 \\ & = k^2 + 2k +2-1 \\ \\ & = k^2 + 2k + 1\end{align}$

Hence, given the inductive hypothesis, it follows that $(k+1)^2 =\sum_{i=1} ^{k+1} (2i-1)$. Hence, $$n^2 = \sum_{i=1}^{n} (2i-1)$$

Jordan Glen
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  • Could I just ask, how you got $k^2 + 2(k+1)-1$ – RandomMath Apr 26 '15 at 14:29
  • hmm I probably miss something but RHS should be $(k+1)(k+2)-(k+1)=(k+1)^2$ – Alex Apr 26 '15 at 14:32
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    @RandomMath $$\sum_{i = 0}^{k + 1} (2i - 1) = \sum_{i = 0}^{k} (2i - 1) + (2 (k + 1) - 1) = k^{2} + (2 (k + 1) - 1)$$ – Matthew Cassell Apr 26 '15 at 14:37
  • As Matteo notes, the sum includes $$\sum_{i = 1}^{k + 1} (2i - 1) = \underbrace{\Big(\sum_{i = 1}^{k} (2i - 1)\Big)}{\large = k^2} +[ (2 (k + 1) - 1)] = \underbrace{k^{2}}{\text{by IH}} + [2 (k + 1) - 1)]$$ – Jordan Glen Apr 26 '15 at 14:40