Binomials and Products

Date: 6/21/96 at 14:0:30
From: Anonymous
Subject: Binomials and Products

In Brazil, we call a Binomial of Newton a number that is represented 
by   /n\ , and that is equal to n!/k!*(n-k)!. 
     \k/      
                                
If S = 1*2*3 + 2*3*4 + 3*4*5 + 4*5*6+...+48*49*50, how can I represent 
the answer of S by using a Binomial of Newton?

I have already noticed that the first term (1*2*3) is equal to 3!/0!, 
and that, therefore, S can be also represented by
S=3!/0! + 4!/1! + 5!/2!+...+50!/47!, in an order that resembles
n!/k!*(n-k)!, but that really is n!/(n-k)!. 

A colleague of mine has told me that I could solve this problem by
trying to relate it to the columns of the Pascal's Triangle, and I 
think he is right, although I cannot figure it out. Could you help me?

P.S. I hope my explanation is not confusing, and sorry if it is. 
I am using / \ as a sign for parentheses.
           \ /

Date: 6/21/96 at 15:50:28
From: Doctor Anthony
Subject: Re: Binomials and Products

You can write the series as:

S = 3! + 4!/1! + 5!/2! + 6!/3! + 7!/4! + .... + 50!/47!

  = 3!{1 + 4!/(3!.1!) + 5!/(3!.2!) + 6!/(3!.3!) + .. + 50!/(3!.47!)}

  = 3!{1 + 4C3 + 5C3 + 6C3 + ...+ 50C3}

  = 3!{4C4 + 4C3 + 5C3 + 6C3 + ... + 50C3}

Now use the fact that nCr + nC(r+1) = (n+1)C(r+1)

  = 3!{(4C4+4C3) + 5C3 + 6C3 + .   + 50C3}

  = 3!{(5C4 + 5C3) + 6C3 + ... + 50C3}

  = 3!{(6C4 + 6C3) + 7C3 + ... + 50C3}

We continue this process combining pairs of terms as shown until we 
arrive at:

S = 3!{50C4 + 50C3}

S = 3!*51C4     

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/