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I'm going through Calculus by Spivak and one of the questions is to find a formula for a summation.

$$\sum_{i=1}^n (2i-1) = 1+3+5+...+(2n-1).$$

I got the correct answer, $n^2$, but did it an alternative method to the answer in the book and wanted some clarity on the working out.

Answer in the book:

$$\begin{align}\sum_{i=1}^n (2i-1) &= 1+3+5+...+(2n-1) \\ &=1+2+3+...+2n-2(1+...+n)\\ &= \frac{(2n)(2n+1)}{2} -n(n+1)\\ &=n^2. \end{align}$$

Why was this approach chosen, from the first to the second line, to me it was not obvious to go down the route of 'adding' in the missing terms, and then subtracting them at the end. Is this a common approach?

I just worked out the two summations, $\sum_{i=1}^n (2i)$ and $\sum_{i=1}^n 1$ added them together and simplified the answer.

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    $\begingroup$ I don't know if there's any obvious reason. You'd have to ask the solution writer. $\endgroup$ – Matthew Leingang Nov 6 '17 at 17:40
  • $\begingroup$ "Is this a common approach?": all tricks are allowed ! $\endgroup$ – Yves Daoust Nov 6 '17 at 17:45
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we have $$\sum_{i=1}^n2i-1=2\sum_{i=1}^n i-\sum_{i=1}^1=2\frac{n(n+1)}{2}-n=...$$ for your second Problem write $$\sum_{i=1}^n(4i^2-4i+1)=4\sum_{i=1}^ni^2-4\sum_{i=1}^ni+\sum_{i=1}^n1$$ can you finish this? HINT: $$\sum_{i=1}^ni^2=\frac{1}{6}n(n+1)(2n+1)$$

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  • $\begingroup$ That is the way I did it. Just wondering why the way noted in the book was taken, as it was not immediately obvious to me to consider the approach. $\endgroup$ – Gurjinder Nov 6 '17 at 17:46
  • $\begingroup$ ok the result is $n^2$, is the Problem solved now? $\endgroup$ – Dr. Sonnhard Graubner Nov 6 '17 at 17:47
  • $\begingroup$ Umm, I guess, the second part is finding the formula for $\sum_{i=1}^n (2i-1)^2 = 1^2+3^2+5^2+...+(2n-1)^2.$ Would there it be possible to take the approach in the question I submitted, rather than the way you have noted? $\endgroup$ – Gurjinder Nov 6 '17 at 17:52
  • $\begingroup$ have you solved it after my hint above? $\endgroup$ – Dr. Sonnhard Graubner Nov 6 '17 at 18:10
  • $\begingroup$ Yeah, thank, but not the second one... $\endgroup$ – Gurjinder Nov 6 '17 at 18:10

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