Finding Binomial Co-efficient from pronumerals


by ukee1593
Tags: binomial coeff, expansions, powers
ukee1593
ukee1593 is offline
#1
Feb10-12, 08:05 AM
P: 2
1. The problem statement, all variables and given/known data

I'm asked to find (a/b) in the simplest form if the co-efficient of x^8 is zero in the expansion of:

(1 + x)(a - bx)^12


2. Relevant equations

Binomial expansion formula ... (a + b)^n = Sum of r --> n (r = 0) (nCr)(a^(n-r) * b^r



3. The attempt at a solution

I figured that x^8 could be achieved from two possible situations ...
either 1 * the expansion of (a - bx)^12 or x * the expansion of (a - bx)^12

I found the value of r at both these points by looking for the value of r that makes bx^r = x^8 and x*bx^r = x^8. This I found to be r = 7 and r = 8. Then I wrote as:

(12C7) * (a)^5 * (-b)^7 + (12C7) * (a)^4 * (-b)^8 = 0

Then I get stuck as I cannot seem to get values for a and b from this.

Can anyone help me?
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ukee1593
ukee1593 is offline
#2
Feb10-12, 09:38 AM
P: 2
Problem solved ...

First of all, note my error in the last line

(12C7) * (a)^5 * (-b)^7 + (12C7) * (a)^4 * (-b)^8 = 0

Should be:

(12C7) * (a)^5 * (-b)^7 + (12C8) * (a)^4 * (-b)^8 = 0

The negative b on the (-b)^7 comes out the front as it is an odd power ... The other negative cancels out

-(12C7) * (a)^5 * (b)^7 + (12C8) * (a)^4 * (b)^8 = 0

throw the negative section over to the other side of the equals sign ...

(12C8) * (a)^4 * (b)^8 = (12C7) * (a)^5 * (-b)^7

then evaluate

=> (495)*(a^4)*(b^8) = (792)*(a^5)*(b^7)

start cancelling

=> (495)*(b) = (792)*(a)

divide to each side

=> 495/792 = a/b

=> a/b = 5/8

which is correct according to the answers.


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