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Binomial Expansions

Date: 03/14/99 at 12:49:02
From: Jeff Doker
Subject: Strange Binomial Expansions

The problem involves finding the constant term of the expansion of
(x + x^-1)^6 (let us define this binomial expansion as a "strange" 
expansion). I know how to find the constant term of expansions of 
normal linear binomials, (x + 1)^6, because it is always the last term, 
but I do not know exactly how to find the constant term of the 
"strange" expansions, since it will not be the last term.

Using some distribution, I can conclude that for "strange" expansions 
of an even power, the constant is the middle term, and for odd powered 
"strange" expansions there is no constant. In addition, the power of 
each term in the "strange" expansion decreases by 2 on each successive 

I then thought I could figure it all out until a friend showed me an 
even stranger expansion problem:

   (x^2 + x^-1)^6

(let us define this one as "weird"). When I tried the same methods, I 
used in solving the "strange" expansion, I found no patterns at all.  

Now I am in search of some formula that uses the degree of each term in 
the binomial, the exponent on the binomial itself, and some number n 
(which is defined as the degree of the term in the expansion that you 
are looking for). This formula will tell you the number of the term 
that you are looking for. 

For example, say you want to find the squared term of the expansion of 
(x + 1)^3.

You give the formula these numbers:
1 (degree of first term)
0 (degree of second term)
3 (degree of binomial)
2 (degree of term in expansion that you are looking for)

The formula should produce the number 2 as the answer. This formula 
should apply to all "strange," "weird," or any other out-of-the-
ordinary binomials. Is there such a formula?

Date: 03/14/99 at 15:30:10
From: Doctor Anthony
Subject: Re: Strange Binomial Expansions

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

    Take the x^2 outside the bracket and we get

     x^(2n)[1 + x^(-3)]^n

     x^(2n)[1 + C(n,1)x^(-3) + C(n,2)x^(-6) + C(n,3)x^(-9) + ....

If n = 3 then the constant term would be C(3,2)

If n = 4 there would be no constant term.

Generally if you have  (x^p + x^q)^n then you make things easier by 
taking one term outside the bracket:

  x^(np)[1 + x(q-p)]^n

  = x^(np)[1 + C(n,1)x^(q-p) + C(n,2)x^(2(q-p)) + C(n,3)x^(3(q-p)) +..

The constant term would be C(n, r) where  np + r(q - p) = 0  if an 
integer value of r satisfies this equation.

- Doctor Anthony, The Math Forum   
Associated Topics:
High School Polynomials

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