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Expansion of n(n+1)(n+2)...(n+k)

Date: 07/15/2002 at 08:03:20
From: Stephen Spackman
Subject: Expansion of n(n+1)(n+2)...(n+k)

Is the expansion of n(n+1)(n+2)...(n+k) known?  I have a very simple 
formula that equals the expansion of n(n+1)(n+2)...(n+k), but I'm not 
sure if anybody is interested in it, or has any uses for it.

Best wishes,
Stephen Spackman.


Date: 07/19/2002 at 09:45:52
From: Doctor Nitrogen
Subject: Re: Expansion of n(n+1)(n+2)...(n+k)

Hello, Stephen:

This is a good question. Sorry, but the expansion of 

   (1) (n + k)(n + k - 1)......(n + 1)(n)

is already known. It is hard to write it in ASCII form, but it is

       SUMMATION(m = 0 to m = k of )S^(m)_k(x^m)

where  S^(m)_k

is called the Stirling number of the First kind. (Here (m) is a 
superscript and k is a subscript in S^(m)_k). Tables for the 
Stirling numbers exist.

Student of mathematics students have, on their own, often tackled 
problems that were already solved, so you are not alone. Note that (1) 
above is called a "factorial polynomial," NOT to be confused with the 
factorial n!

Why not do some research on these numbers in your spare time? After 
you read this, look at this answer from the Dr. Math archives:

   Stirling Numbers
   http://mathforum.org/library/drmath/view/51550.html 

You will find a short tutorial on Stirling Numbers of the First kind. 
Just remember that the variables you used in your question differ from 
the ones you will find there. 

The variables you used appear in:

            (n + k)(n + k - 1).........(n + 1)n,    

But the same expression at the Dr Math link you visit will have:

            x(x - 1)(x - 2).....(x - n + 1).          

So when you visit that Dr. Math link, you will have to replace the 
variables that were used here. To illustrate, suppose in your 
expression for  

            (n + k)(n + k - 1).........(n + 1)n,  
              

that n + k = 8, n = 5, and k = 3. For this answer, then, your 
expression reads

            (n + k)..........n = 8*7*6.....5 

So when you get to the Dr Math link, just let x = 8, and n = 4, so 
that x - n + 1 = 8 - 4 + 1 = 5 there, so that on the link:

       x(x - 1)....(x - n + 1)= 8*7*...(8 - 4 + 1)

                              = 8*7*.....5.

I hope this answered your question. Return with more math questions 
to the Math Forum/Ask Dr. Math.

- Doctor Nitrogen, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Number Theory

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