3

We have this series of numbers:
$1, 3, 6, 10, 15$
The general term can be described wit: $\frac{r(r + 1)}{2}$ Apparently the following series:
$1, 4, 10, 20, 35$
Can be described with $\frac{r(r + 1)(r + 2)}{6}$ based on the first series.
But I am not clear how this is derived. Can someone please explain?

Henrik supports the community
  • 7,050
Jim
  • 727
  • 1
    The first sequence is the sequence of triangle numbers. $T(n)=\sum\limits_{k=1}^n k$. The second sequence you describe is that of the tetrahedral numbers, $Tet(n)=\sum\limits_{j=1}^n T(j) = \sum\limits_{j=1}^n\sum\limits_{k=1}^j k$. There are a number of ways to see this, geometrically, combinatorially, or algebraically. For example here. – JMoravitz Jun 01 '16 at 20:54

4 Answers4

4

Use finite differences:

For the first series, you have $$\begin{matrix} u_n&0&&1&&3&&6&&10&&15\\ \Delta u_n&&1&&2&&3&&4&&5 \end{matrix} $$ The last line is the arithmetic sequence of natural numbers.

For the second series: $$\begin{matrix} v_n&0&&1&&4&&10&&20&&35\\ \Delta v_n&&1&&3&&6&&10&&15\\ \Delta^2v_n&&&2&&3&&4&&5 \end{matrix} $$ One has to solve for $\Delta v_n=u_n$, and the solution is $v_n=\dfrac{n(n+1)(n+2)}{6}$.

Similarly, the solution to $\Delta w_n=v_n$ is $\;w_n=\dfrac{n(n+1)(n+2)(n+3)}{4!}$, and so on.

Added:

This is connected to the problem of finite differences equations. For any polynomial $p(X)\in \mathbf Q[X]$, we define $\;\Delta P(X)=P(X)-P(X-1)$. A finite difference equation is an equation with an unknown polynomial $U(X)$, given a polynomial $P(X)$: $$\Delta U(X)=P(X).$$ To solve it, one has to use an adapted basis of $ \mathbf Q[X]$, which is made up of the polynomials $$ 1,\; X,\; \frac{X(X+1)}{2}, \frac{X(X+1)(X+2)}{3!},\; \dotsm,\; \frac{X(X+1)\dots(X+r-1)}{r!},\;\dotsm $$ which are characterised by the property that the finite difference of each of them is the previous one.

Bernard
  • 179,256
  • I don't understand One has to solve for Δvn=un,. Could you please elaborate? – Jim Jun 01 '16 at 21:11
  • From the first scheme you can see that $u_{n+1}=u_{n}+n$ (for example $n = 2$: $3 = 1+2$). Writing this out will result in $u(n)=1+2+3+4+\cdots n$. Use the same procedure for the second series $u_{n+1}=u_{n}+k_{n}$, where $k_n$ is the numer in the second row with $k_{n}=k_{n-1}+(n-1)$. First find $k_n$ and plug this into $u_{n+1}$ – MrYouMath Jun 01 '16 at 21:18
  • +1: I would have explained it the same way. But is was rather focusing on the problem of uniqueness. – MrYouMath Jun 01 '16 at 21:23
  • @MrYouMath: What do you mean first find Kn? That would be the triangular number of the rank r-1 (or n-1 for your notation) which would be r(r-1)/2 right? – Jim Jun 01 '16 at 21:38
  • Find the general form ok $k_n=1+2+\cdots+(n-1)$ – MrYouMath Jun 01 '16 at 21:39
  • @Jim: I've added some details on finite difference equations, in a more general context. Hope this makes it clearer. – Bernard Jun 01 '16 at 21:40
  • @MrYouMath:First find kn and plug this into un+1 I know that. It is r(r+1)/2). But if I plug it in un+1 I still don't know un – Jim Jun 02 '16 at 20:31
  • $u_{n+1}=u_n+n(n-1)/2$. This will turn out to be a double sum. – MrYouMath Jun 02 '16 at 20:36
  • @MrYouMath:This is the part I am confused. We define un+1 in terms of un which is a recursive definition and are both undefined and the known term r(r+1)/2. How do we approach this? What do you mean double sum? – Jim Jun 02 '16 at 21:01
  • Just try to calculate few terms in the sequence $u_{n}$ and you should see the pattern. – MrYouMath Jun 02 '16 at 21:12
  • But this is what the OP is about. I have the terms for un and I asking how to derive the generating difference. – Jim Jun 02 '16 at 21:14
  • @Bernard: To be honest the added part is not very helpful for me.It is too abstract for me – Jim Jun 07 '16 at 19:02
0

Note: That from the given series you are not able to derive a unique expression for the general term. There are infinitly many solutions. There are some which are more obvious (see JMoravitz), but there is no mathematical definition for more obvious as far as I know.

E.G. a polynomial of 6th degree can fit all points, polynomial of 7th degree ...., all polynomial of higher degree than 6 can have an appropriate choice of coefficients so that all the given points lie on the polynomial.

MrYouMath
  • 16,174
  • What do you mean that there are infinitely many solution? I thought each series has 1 formula to derive a general term. Is that not so? – Jim Jun 04 '16 at 20:48
  • No if you only have a finite number of points. there are mathematically inifinite solutions. imagine you had two points. How many functions can you draw including these two points? Answer is infinite. You could also consider infinite points for integer values and you would still be able to find infinite functions containing all these points. – MrYouMath Jun 04 '16 at 20:55
  • Is there a specific min number of points that leads us to 1 solution? – Jim Jun 05 '16 at 16:16
  • only if you have infinitely many points on the continous domain. Even if there is only one single value which is not given on the interval, you can create an arbitrary jump for this x and still have a function. – MrYouMath Jun 05 '16 at 19:36
  • I am not sure on the last part. How is it possible to do an arbitrary jump and still be consistent with the formula thus far which is build based on the term's position? Is it some kind of mathematical property I am not aware? It seems difficult to come up with 1 formula let alone adapt it by adding new random terms. – Jim Jun 06 '16 at 20:42
  • Do you recommend some good resource that explains the principles of this to read? – Jim Jun 07 '16 at 19:01
0

Those aren't series, they are starts of sequences.

As always when given the start of a sequence, it can continue in any way so there's infinitely many formulas that will describe give a sequence that starts with the given numbers. We can only try to find a simple formula and hope that was what was though of.

Those sequences are well-known, so a lot of people will be able to just give the formulas you have.

If we assume that they might be generated from polynomials of a degree, we can calculate the differences between consecutive elements. For the first sequence this gives: $$ 2,3,4,5 $$ and if we iterate: $$ 1,1,1 $$ So after two steps we got a constant sequence, so it's a second degree polynomial $an^2+bn+c$.

If you do the same with the second sequence, you'll find it takes three steps until you get a constant sequence so the values comes from a third degree polynomial $an^3+bn^2+cn+d$.

The coefficients can be derived from the differences, or you can plug in a few of the known values and get a set of equations for the coefficients.

Henrik supports the community
  • 7,050
0

$S = 1 + 4 +10 + 20 + 35 \cdots + a_{n-1} + a_n \cdots - 1$

$S = 1 + 4 +10 + 20 + 35 \cdots + a_{n-1} + a_n \cdots - 2$

Subtracting 1 from 2

$0 = 1 + ((4 - 1) + (10 - 4) + (20 - 10) + (35 - 20)+ \cdots + a_{n} - a_{n-1}) - a_n$

$a_n = 1 + ((4 - 1) + (10 - 4) + (20 - 10) + (35 - 20)+ \cdots + a_{n} - a_{n-1})$

$a_n = 1 + 3 + 6 + 10 + 15 + \cdots + (a_{n} - a_{n-1})$

$a_n = \sum_{r = 0}^n \frac{r(r + 1)}{2}$

$a_n = \frac{1}{2} * \sum_{r = 0}^n (r^2 + r)$

$a_n = \frac{1}{2} * (\sum_{r = 0}^n r^2 + \sum_{r = 0}^nr)$

$a_n = \frac{1}{2} * (\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2})$

$a_n = \frac{n(n+1)}{2} * (\frac{(2n+1)}{6} + \frac{1}{2})$

$a_n = \frac{n(n+1)}{2} * (\frac{(2n+1 +3)}{6})$

$a_n = \frac{n(n+1)}{2} * (\frac{(2n+4)}{6})$

$a_n = \frac{n(n+1)}{2} * (\frac{(n+2)}{3})$

$a_n = \frac{n(n+1)(n+2)}{6}$

This way is pretty easy.