# B Solve 3 7 12 18 25 series

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1. Mar 2, 2017

### johann1301h

Hi, im trying to solve this series generally:

the series: 3 7 12 18 25.

i tried using x(n) = 3 + 4n.

2. Mar 2, 2017

### PeroK

I would add $0$ at the beginning and see whether that helps.

3. Mar 2, 2017

### Staff: Mentor

Calculate how much the numbers increase in each step. That should lead to some pattern.

4. Mar 2, 2017

### OmCheeto

Did you try graphing it?
When I graphed your series of numbers, I got a curve.
That implied to me, that "4n", being a linear function, was not correct.

5. Mar 2, 2017

### Ssnow

You must find $f(n)$ such that $a_{n+1}=f(n)+a_{n}$ with $a_{0}=3$, hint: think simple!
Ssnow

6. Mar 6, 2017

### Svein

Most sequences can be resolve using difference (not differential) analysis.

Code (Text):

3
7    4
12    5    1
18    6    1
25    7    1

7. Mar 22, 2017

### DShane_MU

Just stumbled across this page. I'm doing independent math research and the series you are inquiring about is a smaller part of what I am doing. The formula you are looking for is F(n)= n(n+5) / 2 = {3,7,12,18,25,...}. It is similar to the formula for triangular numbers T(n) = n(n+1)/2 = {1,3,6,10,15,...}. Hope this helps!

Regards,
David

8. Mar 24, 2017

### maka89

Difference between two numbers is (2+n), so you can write x(n)= (2+n) + x(n-1) when n > 1.

This is a recursive formula, that also can be written out as:
x(n) = (2+n) + (2+n-1) + (2+n-2) + (2+n-3) + ... + x(1)

Recognizing the pattern as:
$x(n) = x(1) + \sum_{i=2}^n(2+i) = x(1) - 3 + \sum_{i=1}^n(2+i) = \sum_{i=1}^n2+ \sum_{i=1}^ni = 2n+ n(n+1)/2 = n(n+5)/2$

So we find x(n) = n(n+5)/2.

$\sum_{i=1}^nn = n(n+1)/2$ can be found from a table or WolframAlpha(or could also derivated without too much trouble) :)