My previous algebra course did not go over summation at all, and now that I'm in my new course, Discrete Math for Information Technology, we have been introduced to summation. I understand summation notation, but I have no idea how to solve equations that contain them. I have looked around on math.stackexchange.com and it seems everyone has specific equations they want solved. What I'd like is something more along the lines of a crash course in solving equations that contain summations. I'm not sure if this question is specific enough, so I'll also include the problem I'm working on.

"A lumberjack has $4n + 110$ logs in a pile consisting of $n$ layers. Each layer has two more logs than the layer directly above it. If the top layer has six logs, how many layers are there?"

I've worked out that each layer has $6+2(i-1)$ logs, where $i$ represents the current layer. I've put this into summation notation to represent all possible layers and set it equal to $4n +110$, which gives me:

$\displaystyle\sum\limits_{i=1}^n [6+2(i-1)]=4n+110$

I am now at a stage where all I have to do is solve for $n$, but I have no idea how to even start since I've not been taught how to solve equations that contain summations.

  • $\begingroup$ Are you familiar with the technique of mathematical induction? $\endgroup$ – anonymous Jan 31 '13 at 13:44
  • $\begingroup$ No, but it is certainly something I could research. $\endgroup$ – Dave Jan 31 '13 at 13:46
  • $\begingroup$ Read a textbook on it, maybe? You could read Concrete Mathematics(The relevant parts).Though that may be more than you need. $\endgroup$ – Ishan Banerjee Jan 31 '13 at 13:48
  • 2
    $\begingroup$ Simplify the left hand side. You can write it as $\sum_{i=1}^n (4+2i)$. This can be further simplified to $4n+2\sum_{i=1}^n i$. So your equation becomes $\sum_{i=1}^n i =55$. Now you could use the fact that $\sum_{i=1}^n i =(n(n+1)/2)$ (or solve by just computing). $\endgroup$ – David Mitra Jan 31 '13 at 13:49
  • $\begingroup$ @DavidMitra That post is very helpful, and I think I will have a solid grasp on this if you could further explain how you got from $\displaystyle\sum\limits_{i=1}^n (4+2i)$ to $4n + 2\displaystyle\sum\limits_{i=1}^n i$. $\endgroup$ – Dave Jan 31 '13 at 14:07

In your case, here are some results for you to either derive or prove through induction:

$$\sum_{i=1}^n 1 = n$$

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

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

$$\sum_{i=1}^n i^3 = \frac{1}{4}n^2(n+1)^2$$

Note that the summation is linear, so that for a term as you have above, you can break it up into

$$\sum_{i=1}^n 6 + \sum_{i=1}^n 2 i - \sum_{i=1}^n 2 = 6 \sum_{i=1}^n 1 + 2 \sum_{i=1}^n i - 2 \sum_{i=1}^n 1$$

  • $\begingroup$ I'm still working through what you're saying here. It takes me a long time to make sense of things like this since I'm new to them, and there's a bit of math anxiety starting to creep up on me (which is not something I'm used to, either). $\endgroup$ – Dave Jan 31 '13 at 14:13
  • $\begingroup$ That's OK. I wrote this post so that you could see things, term for term, and substitute a simple formula for each sum. At the very least, you will be able to see that a sum merely represents a pattern, and in math, all we do is recognize patterns and solve problems with the,. $\endgroup$ – Ron Gordon Jan 31 '13 at 14:15
  • $\begingroup$ I would point out that you typically only need to know the first two of those formulas (but they are very important!). The third one is rarely needed, depending on the level of your course, and the last one is almost never needed. $\endgroup$ – ferson2020 Jan 31 '13 at 14:30
  • $\begingroup$ @ferson2020: at OP's level, sure. But, at that level, they are also asked to prove such formulas through induction, which I urged him to do. In general, though: I have used all of these more times than I can count; thus I can recite them off the top of my head. $\endgroup$ – Ron Gordon Jan 31 '13 at 14:34
  • $\begingroup$ After pounding into my head what those formulas represent, it turns out that they are far more helpful than I originally thought. I'll go ahead and credit you with the answer on this one. Thanks very much for the assistance. Also, there should be 10 layers. $\endgroup$ – Dave Jan 31 '13 at 14:34

You generally have to simplify the summation so that it is no longer there, then solve as you normally would.

For example, $\displaystyle\sum_{i=1}^n[6 + 2(i-1)] = 6n + 2 \sum_{i=1}^{n-1}i$.

You can simplify further and remove the summation completely; then that is equal to $4n + 110$. From there, you can solve for $n$ as you normally would.

  • $\begingroup$ This seems to be exactly what I'm looking for, but I'd like some more information on why that can be done. I can see what you did, but I have no idea why that can be done. I think there is something about summations that I don't understand yet. $\endgroup$ – Dave Jan 31 '13 at 14:11
  • $\begingroup$ I think all you need to know are the formulas that @rlgordonma has given you in his answer, plus how to pull out constants from a sum. For instance, what I did here was first split up the sum over the addition: $\displaystyle \sum_{i=1}^n[6 + 2(i-1)] = \sum_{i=1}^n 6 + \sum_{i=1}^n 2(i-1)$. Then I pulled out the constant $6$ from the first sum, and used the first formula: $\displaystyle \sum_{i=1}^n 6 = 6 \sum_{i=1}^n 1 = 6n$. $\endgroup$ – ferson2020 Jan 31 '13 at 14:27

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