# Summing up 5+9+13+17

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):

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

• 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? Apr 27, 2021 at 22:27
• $\sum_{k=1}^4 (4k+1)$ because you have an Arithmetical Progression with ratio $4$. Apr 27, 2021 at 22:31
• @JeanMarie Common difference, not ratio (ratio is for geometric progression). Apr 27, 2021 at 22:32
• @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) Apr 27, 2021 at 22:34
• Jean Marie thanks, but how could you know, I mean how did you know the solution? i didn't understand Apr 27, 2021 at 22:41

$$\Sigma_{i=0}^{3}{(5+4i)} = 5+9+13+17$$
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:
$$\Sigma_{k=0}^{n-1}{(a_0+kd)}=\frac{n}{2}(2a_0+(n-1)d)$$
and this works case-specifically ($$n=4$$) as well, obviously.
• @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. Apr 28, 2021 at 0:16