# Solve 3 7 12 18 25 series

• B
Hi, im trying to solve this series generally:

the series: 3 7 12 18 25.

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

PeroK
Homework Helper
Gold Member
Hi, im trying to solve this series generally:

the series: 3 7 12 18 25.

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

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

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

OmCheeto
Gold Member
Hi, im trying to solve this series generally:

the series: 3 7 12 18 25.

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

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.

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

Aufbauwerk 2045
Svein
Most sequences can be resolve using difference (not differential) analysis.

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

Aufbauwerk 2045
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

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) :)

Type out math instead of posting images.

etotheipi
Gold Member
2019 Award
This is not necessarily a better method, since yours is actually probably faster in this case, however any series with a constant second difference is called a quadratic series and can be generalised as such:

##u_{n} = an^{2} + bn + c##

First term: ##u_{1} = a + b + c##
First difference: ##u_{n} - u_{n-1} = 2an + (b-a)##
Second difference: ##(u_{n} - u_{n-1}) - (u_{n-1} - u_{n-2}) = 2a##

So in the case of the sequence ##3,7,12,18,25...## we can write down three relations

##a + b + c = 3##
##3a + b = 4##
##2a = 1##

Which gives ##a = \frac{1}{2}, b = \frac{5}{2}, c= 0##, so

##u_{n} = \frac{1}{2}n^{2} + \frac{5}{2}n##

which is the answer you obtained.

wle
More generally it's easy to check by induction that $$\sum_{n=1}^{a} n (n + 1) (n + 2) \dotsm (n + p - 1) = \frac{a (a + 1) \dotsm (a + p)}{p + 1} \,.$$ The relation can be be seen as the discrete analogue of $$\int_{0}^{a} \mathrm{d}x \, x^{p} = \frac{a^{p+1}}{p + 1}$$ and it can be used to determine the partial sum of any polynomial. For example, $$\begin{eqnarray*} \sum_{n=1}^{a} n^{3} &=& \sum_{n=1}^{a} \bigl( n - 3 n (n + 1) + n (n + 1) (n + 2) \bigr) \\ &=& \frac{1}{2} a (a + 1) - 3 \frac{1}{3} a (a + 1) (a + 2) + \frac{1}{4} a (a + 1) (a + 2) (a + 3) \\ &=&\frac{1}{4} a^{2} (a + 1)^{2} \,. \end{eqnarray*}$$