Out of my own interest I've been practicing finding formula for for sequences and I've been having trouble finding one for the nth term for this sequence.
0,3,8,15,24 ...
Clearly you add 5,7,9,11 ... to the previous number but if anyone had some insight about how to express this in a formula that would be much appreciated.
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$\begingroup$Add $1$ to each term. Do you recognize it now?
$\endgroup$ 5 $\begingroup$Clearly, $a_n-a_{n-1}=2n+1,n\ge1$
Let $a_m=b_m+p+qm+rm^2,a_0=0\implies b_0=-p$
$2n+1=b_n-b_{n-1}+q+r(2n-1)$
Set $2r=2,q-r=1$ so that $b_n=b_{n-1}=\cdots=b_0=?$
$\endgroup$ 4 $\begingroup$Ok, so your Growth Rate is f(x) = 2n + 1 As this is Growth, you have to find the Mother Function F(x) = n squared + n ( + c =n in this case), thats it. So its: n squared + 2n !
$\endgroup$ $\begingroup$A recursive general formula seems to be
$$ n_0 = 0 $$$$ n_k = n_{k-1} + 2k+1 $$
$\endgroup$ $\begingroup$The solution for $ a(n) $ is this$ a(n) = n(n+2) = (n+1)^2 - 1. $
$\endgroup$ $\begingroup$Well obviously $a_n=n^2-1, n\in \mathbb{N}.$
$\endgroup$ $\begingroup$If you look at "really old" numerical analysis books (before about, maybe, 1970), they almost always had a section on (at least) forward differences. You would write down your sequence in a column (maybe with the indices in the column to the left, so you don't get lost), and then in the spaces between the a(i) and to their right, you would write a(i) - a(i-1) = delta(a(i-1)). This is the forward difference. It's a finite difference, the finite analog of the numerator when taking the derivative. If you take the forward difference of the forward difference, that's the second difference. There is one difference from taking derivatives: we could also look at a(i-1) - a(i-2) = del(a(i-1)) = the backwards difference. (I may have my notations backwards. I haven't actually done this stuff since maybe 1985. I mean, with the notation. There's also a central difference, with a minuscule [as opposed to capital] delta, and even off-center differences. People went hog-wild with this stuff before computers, and then it just disappeared in an instant.)
Anyway, you keep taking differences until you get a pattern you recognize. (At this stage, that's probably a constant.) Then you go backward.
So let's say the second difference is 1. Then the first difference (divided by the interval) is n and, just like integrating, we have to add a constant. So if we're working with a sequence (this works just as well for a function), the formula for the first difference is n + C. (We can probably read C off of our table of first differences.) Now we need to know what has a first difference of n? Well, what do you get if you add the first n integers? n(n+1)/2 of course, so that's always the anti-difference of n, and the terms will be n(n+1)/2 + Cn + D.
$\endgroup$ 3 $\begingroup$There are some rules of thumb for sorting out sequences like this.
Some sequences are Arithmetic Progressions: $a_0, a_0+k, a_0+2k,...$
Some sequences are Geometric Progressions: $a_0, a_0+k, a_0+k^2, a_0+k^3, ...$
Some are more general Iterative Processes: $a_{n+1}=a_n+a_{n-1}, a_0=0, a_1=1,$ e.g. $0,1,2,3,5,8,...$
Each of these exhibits different behaviors in their forward differences, the sequence of the difference in their consecutive terms. In arithmetic sequence, this is a constant. In a geometric sequence its a difference of consecutive powers of the common ratio.
Sob often you want to start out calculative difference between consecutive terms. You might want to do this for multiple layers, i.e. calculate the consecutive differences in the sequence of differences.
You might also want to break squences into two sequences before you start taking differences. Sort out the even numbered terms from the odd numbered terms, then perform your difference analysis matching them against the patterns mentioned above.
There are more advanced techniques like Generating Functions and associated Auxiliary Equations but those above are good starting points.
Here's a fun squence to mess around with.
The sum of the first N integers is $\frac{N(N+1)}{2}$
The sum of the squares of the first N integers, $1^2+2^2+...+N^2 = \frac{N(N+1)(2N+1)}{6}$
The sum of cubes is $\frac{N^2(N+1)^2}{4}$
You have a sequence of sequences. What is the formula for adding the fifth powers of the first N inters?
$\endgroup$ $\begingroup$You need to know Arithmetic sequence and Arithmetic series.
let $n_1 = 0$ (it means let's counting from 0)$$a_n = a_1+\sum_{k=0}^n b_k$$
$b_k = a_1 + d(n)$ ($b_k$ is an arithmetic sequence of differences)
$ = 3+2(n)$
$ = 2n +3$
$ a_n = 0 + \sum_{k=0}^n 2n + \sum_{k=0}^n 3$
$= 2\sum_{k=0}^n n + \sum_{k=0}^n 3$
$= 2(n(n+1)/2)+n$
$= n^2+n+n$
$= n^2+2n$
$a_1=(0)^2+2(0)=0$
$a_2=(1)^2+2(1)=3$
$a_3=(2)^2+2(2)=8$
$a_4=(3)^2+2(3)=15$
$a_5=(4)^2+2(4)=24$
$\endgroup$ $\begingroup$$$x_{n+1}=x_n+(2n+1) \text{ , with } x_1=0$$ Now check this $x_2=0+(2.1+1)=3, x_3=3+(2.2+1)=8,x_4=8+(2.3+1)=15\text{ and so on.}$
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