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Expanding Binomials and Pascal's Triangle

Date: 06/07/98 at 22:12:57
From: Tamara Hatch
Subject: Binomials and Pascal's triangle

                             5*4              5*4*3
(a+2b)^5 = a^5 + 5*a^4(2b) + --- a^3 (2b)^2 + ----- a^2 (2b)^3
                             1*2              1*2*3
           + -------- a (2b)^4 + (2b)^5

        = a^5 + 10 a^4 b + 40 a^3 b^2 + 80 a^2 b^3 + 80 a b^4 + 32 b^5

One of my friends got this answer from you, and I was wondering if you 
could explain more in words about expanding the binomal (a+2b)^5.

Date: 06/09/98 at 08:42:49
From: Doctor Jerry
Subject: Re: Binomials and Pascal's triangle

Hi Tamara,

To expand any binomial of this kind, first write a^5, the first term.  
The coefficient of this term is 1 and it is the first term. The 
coefficient of the second term (which has a^4 and (2b)^1 in it) is the 
exponent of a in the first term times the existing coefficient (1), 
divided by the number of the term. So, the coefficient of the second 
term is 5*1/1 = 5. 

Okay, so now we have:

   a^5 + 5*a^4*(2b)^1

To get the coefficient of the third term, apply the same rule. Take 
the exponent of a in the second term, multiply by the coeffcient of 
the second term and divide by the number of the term. So, 4*5/2 = 10.  

Now we have:

   a^5 + 5*a^4*(2b)^1 + 10*a^3*(2b)^2

Next, 3*10/3 = 10:

   a^5 + 5*a^4*(2b)^1 + 10*a^3*(2b)^2 + 10*a^2*(2b)^3

Next, 2*10/4 = 5, and so on.

You can see that we have generated 1 5 10 5 1, a line from Pascal's 
triangle. This gives the same results you listed in your message.

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

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