How did we derive this general term for the series?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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

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

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.