Expanding Binomials and Pascal's TriangleDate: 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 5*4*3*2 + -------- a (2b)^4 + (2b)^5 1*2*3*4 = 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/ |
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