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My vocabulary in math is lacking quite a lot, so please forgive me if my question is not sufficiently accurate or needlessly verbose. I tried very hard to get the latex flowing, at least that's one thing I got going.

In the syllabus from which I'm currently studying, the proof of this equation is something I have difficulty grasping (this is a quick example that's supposed to be self-explanatory *desperate face*):

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

Now, this is the proof in the syllabus:

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

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

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

I understand the reasoning, 1 being the minimum value and valid allows for rewriting the equation for k+1, and if everything for the k+1 version is valid, along with the minimum of k being valid, everything is valid.

However, why isn't the first step done like this? (subtraction, isn't valid but seems logical to me atm)

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

Also, could someone please point me in the right direction on how to accomplish this conversion:

$$k^2+2k+1 = (k+1)^2$$

Thank you in advance for helping me find the sources on how to understand everything until page 18 (out of 180 ^_^)

Ian Mateus
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Chris B.
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  • It's $\sum_{i=1}^k (2i-1) + (2(k+1)-1)$. – Daniel Fischer Dec 15 '13 at 19:25
  • it's 2k+1, because you have (2(k+1)-1) for (k+1), and the convergences is done from the induction theorem. – randomname Dec 15 '13 at 19:27
  • And $(k+1)^2=(k+1)(k+1)=k\cdot k+k\cdot1+1\cdot k+1\cdot1=k^2+2k+1$. (This is a special case of the binomial formula $(a+b)^2=a^2+2ab+b^2$.) – Carsten S Dec 15 '13 at 19:31
  • Thanks a 1000, the first step makes perfect sense now. Can't believe I missed it. – Chris B. Dec 15 '13 at 19:34
  • Thanks a lot for the second part as well Carsten. – Chris B. Dec 15 '13 at 19:41
  • How about this: compute: $k^2-(k-1)^2$, you get: $k^2-(k^2-2k+1)=2k-1$ . So , you start with a square --1=1^2-- and then you add the differences to consecutive squares. So you start with a square, and then you add the difference to the next square, etc. – user99680 Dec 15 '13 at 20:17
  • Hi, thanks for showing a different way on how to solve it, really appreciated. – Chris B. Dec 16 '13 at 19:54

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