I have a summation and I want to be able to find the sum for given $n$ without having to go through $1,\dots,n$.

$$\sum_{x=1}^{n - 1}x+300\cdot2^{x/7}$$

It's been awhile since I've done summations and I can't figure this one out. Also, this isn't for homework if that makes a difference.


  • 1
    $\begingroup$ Solve for $x$ does not seem like the right question, since $x$ is a dummy variable of summation. Maybe you want the sum in terms of $n$. We can split into two parts, an arithmetic series and a finite geometric series. There are simple formulas for each part. $\endgroup$ – André Nicolas Jan 9 '15 at 3:47
  • $\begingroup$ What do you want to solve for $x$ ? What is the equation to solve ? And, as André Nicolas pointed out, is $x$ the index for summation ? Please clarify. $\endgroup$ – Claude Leibovici Jan 9 '15 at 3:51
  • $\begingroup$ @AndréNicolas You're right, I worded it incorrectly. I'm not very well-versed in math, so I apologize. $\endgroup$ – Stripies Jan 9 '15 at 3:52
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    $\begingroup$ This is just nonsense, I do not think this is a real question. $\endgroup$ – VividD Jan 9 '15 at 3:52
  • $\begingroup$ @VividD: After a while, what is intended will become clear. $\endgroup$ – André Nicolas Jan 9 '15 at 3:56

Might as well give an answer. Note that we can split this sum into the following two sums: $$\sum_{x=1}^{n-1} x + 300 \cdot 2^{x/7} = \sum_{x=1}^{n-1} x + \sum_{x=1}^{n-1} 300 \cdot 2^{x/7}.$$

Given that $$\sum_{x=1}^n x = \frac{n(n+1)}{2}$$ we know that $$\sum_{x=1}^{n-1} x = \frac{n(n+1)}{2} - n = \frac{n(n-1)}{2}.$$ We're halfway done! Turning our attention to the second sum, we have that \begin{align*} \sum_{x=1}^{n-1} 300 \cdot 2^{x/7} & = 300 \sum_{x=1}^{n-1} 2^{x/7} \\ & = 300 \left(\frac{2^{n/7} - 2^{1/7}}{2^{1/7} - 1}\right) \end{align*} by the formula for a finite geometric series.

Therefore, the total sum is $$\sum_{x=1}^{n-1} x + 300 \cdot 2^{x/7} = \frac{n(n-1)}{2} + 300\left(\frac{2^{n/7} - \sqrt[7]{2}}{\sqrt[7]{2} - 1}\right).$$ Plug in $n$ and you will have your solution.

  • $\begingroup$ Thanks, that is what I meant, despite not saying it correctly. $\endgroup$ – Stripies Jan 9 '15 at 5:48
  • $\begingroup$ No problem! In the future, try fiddling around with the sum a bit, using whatever references necessary (the Wikipedia article on summation would probably suffice in this case), and see how far you can get. Then we can see where you got stuck and help fill in any gaps in your knowledge. If we don't see any work put forth by you, it looks like you aren't willing to put in any effort to get the answer, which suggests that you just want us to give you the end result without bothering to learn the process to get there. $\endgroup$ – Clark Zinzow Jan 9 '15 at 6:14
  • $\begingroup$ Even if you aren't well-versed in math, even if you don't care about the process used to get an answer to a question like this, show us that you're willing to take a couple minutes of your time to learn how to solve at least a portion of the problem, and this community will be a lot receptive towards your question. That 2-3 minutes of effort could have saved yourself 10-15 minutes of waiting if I hadn't answered the question, since I'm sure that many others would have passed on this question once they saw that you didn't show any of your work towards a solution. $\endgroup$ – Clark Zinzow Jan 9 '15 at 6:16
  • $\begingroup$ Food for thought I suppose! $\endgroup$ – Clark Zinzow Jan 9 '15 at 6:19

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