0
$\begingroup$

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.

$\endgroup$
4
  • 2
    $\begingroup$ I don't know if there's any obvious reason. You'd have to ask the solution writer. $\endgroup$ Nov 6, 2017 at 17:40
  • 1
    $\begingroup$ "Is this a common approach?": all tricks are allowed ! $\endgroup$
    – user65203
    Nov 6, 2017 at 17:45
  • $\begingroup$ This solution doesn't appear in the solution manual of the book. $\endgroup$ Aug 12, 2022 at 21:44
  • $\begingroup$ Yes, it is the obvious approach, for problems of this nature. An alternative approach is to recognize that $~1 + 3 = 2^2, ~1 + 3 + 5 = 3^2,~$ and then form a hypothesis, and then prove it by induction, using the fact that $~(k+1)^2 - k^2 = 2k+1.$ $\endgroup$ Sep 12, 2023 at 16:24

3 Answers 3

1
$\begingroup$

The method they use is pretty common in calculating sums. Your method is the more obvious one and what I would have used in this question. However, at times the book's method is superior. For example, I can't find a way of calculating $$\sum_{k=0}^\infty\frac{1}{(2k+1)^2}$$using your method. Using their method, we would rewrite this sum as $$\sum_{k=1}^\infty\frac1{k^2}-\sum_{k=1}^\infty\frac1{(2k)^2}=\sum_{k=1}^\infty\frac1{k^2}-\frac14\sum_{k=1}^\infty\frac1{k^2}=\frac34\sum_{k=1}^\infty\frac1{k^2}=\frac34\times\frac{\pi^2}6=\frac{\pi^2}8$$I think the textbook's authors didn't use the method you used to prepare you for more difficult sums in the future. I don't have any other idea why they would use it in this problem.

$\endgroup$
0
$\begingroup$

I think they do it in books because primitive techniques (tricks) are simple and common.

Here is the trick:

$\small\begin{array}{c} S=&1&+&3&+&...&+&(2n-3)&+&(2n-1)\\ S=&(2n-1)&+&(2n-3)&+&...&+&3&+&1\\ 2S=&2n&+&2n&+&...&+&2n&+&2n&\text{(n times)}\\ 2S=&n(2n)\\ S=&n^2 \end{array}$

This idea is too old. We learned it from our teachers. You may see A. Nesin's nice illustration of this idea in youtube shorts.

$\endgroup$
-1
$\begingroup$

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: $$\sum_{i=1}^ni^2=\frac{1}{6}n(n+1)(2n+1)$$

$\endgroup$
6
  • $\begingroup$ 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, 2017 at 17:46
  • $\begingroup$ ok the result is $n^2$, is the Problem solved now? $\endgroup$ Nov 6, 2017 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, 2017 at 17:52
  • $\begingroup$ have you solved it after my hint above? $\endgroup$ Nov 6, 2017 at 18:10
  • $\begingroup$ Yeah, thank, but not the second one... $\endgroup$
    – Gurjinder
    Nov 6, 2017 at 18:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .