Summing up 5+9+13+17 using sigma symbol ($\sum$) I tried this, but I couldn't, but I think 4 is added every time.

My answer was (I think it's wrong):

$$ \sum_{i=5}^{17} (i+4) $$

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    $\begingroup$ Welcome to MathSE. Normally we need to see some effort on your part to answer your question and find precisely where you are confused. Can you more clearly describe the question, and describe your answer? $\endgroup$
    – Pavan C.
    Apr 27, 2021 at 22:27
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    $\begingroup$ $\sum_{k=1}^4 (4k+1)$ because you have an Arithmetical Progression with ratio $4$. $\endgroup$
    – Jean Marie
    Apr 27, 2021 at 22:31
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    $\begingroup$ @JeanMarie Common difference, not ratio (ratio is for geometric progression). $\endgroup$
    – Deepak
    Apr 27, 2021 at 22:32
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    $\begingroup$ @Deepak Thanks for the right term I didn't know in english (in french we use the same term "raison = ratio" for AP and GP) $\endgroup$
    – Jean Marie
    Apr 27, 2021 at 22:34
  • $\begingroup$ Jean Marie thanks, but how could you know, I mean how did you know the solution? i didn't understand $\endgroup$ Apr 27, 2021 at 22:41

1 Answer 1


Depends; if you have simply a finite set of numbers (which I think is the case here), you can, in an elementary way, express it as:

$$\Sigma_{i=0}^{3}{(5+4i)} = 5+9+13+17$$

But in a more general case, as this is obviously an arithmetic sequence with $a_0=5$ and $d=4$, you can use the summation formula for the arithmetic sequence:


and this works case-specifically ($n=4$) as well, obviously.

  • $\begingroup$ How did you find the first answer ? $\endgroup$ Apr 28, 2021 at 0:14
  • $\begingroup$ @aymenaymen An arithmetic sequence is a sequence with a common difference; in this case, your set of numbers is a sequence with a constant difference $4$. Then, you plug in the initial number $a_0$ and the common difference $d$ in the general summation formula, which in this case are $5$ and $4$ respectively. $\endgroup$
    – raven
    Apr 28, 2021 at 0:16

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