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A Different Binomial Expansion Theorem


Date: 07/23/98 at 09:22:44
From: Galip Adalan
Subject: A new theorem about the famous binomial opening

Dear Dr. Math:

A few weeks ago, one of the students in a class that I teach asked me
a brief and specific question about the famous binomial theorem. The 
student claims there is a completely different and and interesting 
form of the opening, which may be stated as following:

(a+b)^n = (a^n + b^n) - T(n,1)*(a+b)^(n-2)*(a*b) + 
          T(n,2)*(a+b)^(n-4)*(a*b)^2 - T(n,3)*(a+b)^(n-6)*(a*b)^4 
          + ...

An alternating series, in which T(n,r) is a completely new coefficient 
of expansion. 

The student wants me to prove or state the theorem in the classroom.
I haven't been able to locate the theorem. I'll owe my gratitude to 
you if you handle the case for me. Thanks in advance.

Galip Adalan


Date: 07/24/98 at 10:47:18
From: Doctor Rob
Subject: Re: A new theorem about the famous binomial opening

The correct formula is:

   (a+b)^n = a^n + b^n - Sum (-1)^r * T(n,r) * (a+b)^(n-2*r) * (a*b)^r

where the sum runs over 1 <= r <= n/2. It is more usually given in the
following variant form:

   a^n + b^n = Sum (-1)^r * T(n,r) * (a+b)^(n-2*r) * (a*b)^r

where the sum runs over 0 <= r <= n/2, and n >= 1.

The initial conditions and recursion that the T's satisfy is:

   T(n,0) = 1 for n >= 1
   T(n,r) = 0 if n < 2*r
   T(n+1,r) = T(n,r) + T(n-1,r-1)

This allows you to compute the values of T from the values with smaller 
n's and smaller or equal r's. First I would fix r = 0, and write down 
the values of T(n,0) = 1 in the first column. Then I would fill in 0's 
according to the second initial condition. Then I could compute T(n,1) 
for n = 2, 3, 4, .... Then I could compute T(n,2) for n = 4, 5, 6, ..., 
and so on.

You can express the T(n,r) in terms of the binomial coefficients as
follows:

   T(n,0) = C(n+1,0)
   T(n,1) = C(n,1)
   T(n,2) = C(n-1,2) - C(n-3,0)
   T(n,3) = C(n-2,3) - C(n-4,1)
   T(n,4) = C(n-3,4) - C(n-5,2)

and so on. In general:

   T(n,r) = C(n-r+1,r) - C(n-r-1,r-2)

You can verify that this expression for T(n,r) does satisfy the above 
recursion and initial conditions, and so must be equal to T(n,r).

Using this, you can now prove the second formula given above using the
Principle of Mathematical Induction, where you assume the statement
true for n, then prove it for n+1. This proof is similar to the proof
usually given for the Binomial Theorem.

If this information is not sufficient to finish the proof, write back
again and we can supply more detail.

- Doctor Rob, The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
    
Associated Topics:
High School Polynomials

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