What is the formula to determine the values 2000, 3720, 5361, 6924,
8410, 9820 .....?

 Here's the finite differences triangle:
T2    2000    3720    5361    6924    8410    9820    ...
T1        1720     1641  1563     1486     1410   ...
CN            80         79        78        77    .....
                  1           1           1     ....

I am not a mathematician but a programmer. It is difficult to express
this algorithm into mathematical terms.
Simple Basic program do describe the behavior of the algorithm.
.
 T1=1800 :CN =81 : T2=2000 :X1 =0
   PRINT "X1=";X1;"   CN=";CN; "   T1="; T1 ;"   T2=";T2
             FOR X1  = 1 TO 26
                  CN=CN-1
                  T1 = T1-CN
                  T2= T1+T2
    PRINT "X1=";X1;"   CN=";CN; "   T1="; T1 ;"   T2=";T2
              NEXT X1

Note: X1 is only there to structure the loop. Not part of any value.

How do you calculate the formula?

In this case, what do you call the different parts of the Pascal Triangle?

Any web references which treat of the same problem, would be appreciated?

Note:  The value of the variables CN & T1 have been sized down.
The original value of  CN > 100 000000 and T1>800 000000 .

I need a formula because going through the loops takes too many
iterations +  enormous computing time.

If you do not understand what I am talking about, please ask me . I'll
be too happy to explain.

gnh888

You've written your question much more clearly now. I'm not going to
provide the answer for what looks like a homework problem, but I will
give you one way to proceed.

Each row is a polynomial starting with zeroth order at the bottom and
3rd order at the top. There are four coefficients in a cubic
polynomial and you have four rows of results. You can solve for the
coefficients, or follow this procedure.

Row 0: R0(N) = 1                       (Zeroth order polynomial)
Row 1: R1(N) = 80 - N                  (First order polynomial)
Row 2: R2(N) = Use the formula for summing an arithmetic series that I gave
   in answer to your earlier question. It yields a quadratic in N.
Row 3: R3(N) = Use a similar formula for summing a sequence of
   squares. Use this in conjunction with the formula for summing an
   arithmetic sequence to yield your result.

Note that the sum from one to N of N^2 is: N(N+1)(2N+1)/6

guyyy,

If you don't mind, I would like to ask four questions.   (You don't
have to answer any of them.)

1) Two people think that this is a homework problem. Is this for a
discrete mathematics course of some type? For a programming course?

2) How did you go about choosing the value of 26 for the end of the
loop? (It gives a local maximum for T2.) Was there a reason for not
going past 26?

3) How far did you get in using either of kottekoe's two methods?

4) In your course, have you covered polynomials in relation to Pascal
triangles? Were tetrahedral numbers covered?

I actually feel it isn't a homework problem, since in that case he
probably would have left the original numbers intact - and the first
question would probably have been written clearly. But I'll provide my
own set of numbers and skip some parts, just in case.

T2  1000000  1220000  1592300  2116899  2793796  3622990
T1       220000   372300   524599   676897   829194
CN            152300   152299   152298   152297

As kottekoe said, the values for T2 will follow a cubic equation,
which means there are four coefficients; and you can approach it
either by direct substitution of known values, or by working it out
from the summation formula.

By substitution: We know T2(n) = An^3 + Bn^2 + Cn + D for some A, B, C and D.
T2(1) = 1000000 ==> A.1 + B.1 + C.1 + D.1 = 1000000
T2(2) = 1220000 ==> A.(2^3) + B.(2^2) + C.2 + D = 1220000
T2(3) = 1592300 ==> A.(3^3) + B.(3^2) + C.3 + D = 1592300
T2(4) = 2116899 ==> A.(4^3) + B.(4^2) + C.4 + D = 2116899

So we get four equations:
  A +   B +  C + D = 1000000
 8A +  4B + 2C + D = 1220000
27A +  9B + 3C + D = 1592300
64A + 16B + 4C + D = 2116899

These equations can be solved simultaneously, either directly or using
matrices, if you remember how to do that. For these values I get
A = -1/6
B = 76151
C = -50711/6
D = 932301

which you can check by putting them in the formula and calculating the
first four (or more) values.

The other method kottekoe suggested is to find formulae successively
for CN, T1 and T2. CN(n) is obviously 152301 - n in this case (I'll
set n=1 for the first element in each line).

Now T1(n) = 220000 + CN(1) + CN(2) + ... + CN(n-1)
 = 220000 + (152301 - 1) + (152301 - 2) + ... + (152301 - n + 1)
 = 220000 + (152301 + ... + 152301) - (1 + 2 + ... + (n-1))
[where there are (n-1) terms in each of the two sequences]
 = 220000 + (n-1)*152301 - (n-1)(n)/2 using the formula for the sum of
the first N integers, which is N(N+1)/2 - here N = n-1.
 = 220000-152301 + 152301 n - n^2 / 2 + n/2
 = 67699 + (304603/2)n - (1/2)n^2

Then we do the same thing for T2. I'll approach it a bit differently
this time - instead of expanding and regrouping, we'll consider it as
a sum of the formula for T1. This is both faster and cleaner...

T2(n) = 1000000 + T1(1) + T1(2) + ... + T1(n-1)
 = 1000000 + sum(i=1..n-1, 67699 + (304603/2)i - (1/2)i^2)
 = 1000000 + 67699 (n-1) + (304603/2)(n-1)(n)/2 - (1/2)(n-1)(n)(2n-1)/6
using the formula from the last step and the formula kottekoe gave for
the sum of the first N squares.

If you simplify this, you get the same formula that we got before (as
you'd expect).

Personally I find substitution and solving the equations to be an
easier way to do it, but that is just a personal preference - you
probably get a better feel for the process by doing it the other way.
But really it comes down to which approach you're most comfortable
with. And remember to check your answers by calculating the first
several terms!

guyyy,

This comment may be more for my benefit than yours. (Prior to your
problem, I would have sworn - and probably will still swear - that
I've never seen this type of problem before; but, then in one of the
sites mentioned below, the author says he learned how to do this in
the 7th to the 9th grade. Of course, he/she turns out to be a
mathematician.)

You asked for a web site having another example. I think that I have
found some that may be what you want. I was aided greatly by the
comments of others and by terms you used.

Here is a site that does a good job of explaining the problem and how to tackle it:
http://mathforum.org/library/drmath/view/53223.html

A proof that the polynomial method works is given at:
http://mathforum.org/library/drmath/view/56953.html

Here is another site that goes through an example, but I prefer
work-through in the first site listed above.
http://www.math.ilstu.edu/day/courses/old/305/contentfinitedifferences.html

The reason I mention this site is that the sequence used are
tetrahedral numbers (more on this below).

The key to finding these sites were two terms you used. The search I used was 

"finite differences" polynomial.

I've heard of the term "finite differences" before, but I know that
I've never had a course in it.

Anyway, kottekoe's methods were also helpful to me in another way;
they led me to search on

"pascal's triangle" square

which led to square numbers, then to tetrahedral numbers. I recognized
these as the way to finish the incomplete solution I had in your first
post. This gives a third way to solve the problem. Here is the formula
I get when the formula for tetrahedral numbers is substituted for the
series I had in my incomplete solution. (This formula has been
adjusted to account for the use of both 2000 as the base number and
1800 as the starting number for the added part.)

200 + 1800*n - 80*n *(n-1)/2 +n(n-1)(n-2)/6

This is mathematically identical to what I get by the polynomial
method listed by kottekoe (and in the sites listed above).

I never was able to discern the "squares" in the second method listed
by kottekoe (probably something obvious that I missed), but that
forced me to searching that led me to the tetrahedral number solution.

The third differences are constant. So it is a cubic. Solving for the
coefficients   I get the nth term is 1/6*n^3 - 40*n^2 + 10559n/6 +
2000, where 2000 is the 0th term.

berkeleychocolate-ga did not explicitly mention that the two solutions

200 + 1800*n - 80*n *(n-1)/2 +n(n-1)(n-2)/6

and

1/6*N^3 - 40*N^2 + 10559N/6 + 2000

give the same results (in case anyone's wondering).

The forms of the formulas reflect their origins and their differences
in numbering. The first formula uses n=1 in referring to the first
result in the series (2000) while the second formula uses N=0 in
referring to the first result in the series. The formulas can be
interconverted by using n=N+1 (or N=n-1).