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Binomial Expansion problem

  • Thread starter Peter G.
  • Start date
  • #1
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Hi,

When (1+ax)n is expanded as a series in expanding powers of x, the first three terms are 1 - 8x + 30x2

Calculate the values of a and n.

So, I think we need simultaneous equations and I managed to build the first one:

a*n = 8

My problem is however, that to construct the second equation I need to transform nCr in to something I can work with. I know it is something around

n(n-1)(n-2) / 2!​

But I am not sure and I don't know how to work with these. Could anyone please help me?

Thanks!
Peter G.
 

Answers and Replies

  • #2
33,630
5,288
Hi,

When (1+ax)n is expanded as a series in expanding powers of x, the first three terms are 1 - 8x + 30x2

Calculate the values of a and n.

So, I think we need simultaneous equations and I managed to build the first one:

a*n = 8

My problem is however, that to construct the second equation I need to transform nCr in to something I can work with. I know it is something around

n(n-1)(n-2) / 2!​

But I am not sure and I don't know how to work with these. Could anyone please help me?

Thanks!
Peter G.
The coefficient of the x2 term will be n(n-1)(n-2) / 2! times a2, and you know that this coefficient has to be 30.

Your first equation is really a*n = - 8. Since both a and n have to be integers, there aren't a lot of choices for either one. You don't need to try n = 1, 3, 5, 6, or 7.
 
  • #3
442
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So when I have the two equations I don't need to do simultaneous? Only trial and error?
 
  • #4
33,630
5,288
You could do the problem either way, but I think that educated trial and error would be much quicker.
 
  • #5
442
0
Ok, I think the trial and error is better in this case because the simultaneous is quite complicated...
 
  • #6
33,630
5,288
The coefficient of the x2 term will be n(n-1)(n-2) / 2! times a2, and you know that this coefficient has to be 30.

Your first equation is really a*n = - 8. Since both a and n have to be integers, there aren't a lot of choices for either one. You don't need to try n = 1, 3, 5, 6, or 7.
I didn't check your work, and now I see an error. The first equation is an = -8, but the second one should be n(n - 1)/2 * a2 = 30, not n(n-1)(n-2) / 2! * a2.
 
Last edited:
  • #7
442
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Oh, ok, so when it is n N 3 for example, it should be n(n-1)(n-2)?
 
  • #8
33,630
5,288
Do you mean nC3? That would be n!/[3! (n - 3)!] = [n(n - 1)(n - 2)]/3!
 
  • #9
442
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Yeah, sorry, that was what I meant. Thanks!
 
  • #10
33,630
5,288
This seems to be harder than it should be - are you sure you have the right coefficients?
 
  • #11
442
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Yes, the answers the book has are: n = 16 and a = -0.5
 
  • #12
33,630
5,288
Yeah, those work.

It still comes down to solving the system
an = -8
n(n - 1)a2 = 60

The second equation comes from n(n - 1)/2 * a2 = 30

From the first equation, a = -8/n.

Substitute this into the second equation to get
n(n - 1) 64/n2 = 60

or
64 (n2 - n)/n2 = 60

This is the same as 64 (1 - 1/n) = 60, or
(n - 1)/n = 15/16, whose solution is n = 16.
 

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