How Do You Solve a Binomial Expansion Problem with Given Series Terms?

[Peter G.]

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 into 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.

 

[Mark44]

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 into 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.

 

[Peter G.]

So when I have the two equations I don't need to do simultaneous? Only trial and error?

 

[Mark44]

You could do the problem either way, but I think that educated trial and error would be much quicker.

 

[Peter G.]

Ok, I think the trial and error is better in this case because the simultaneous is quite complicated...

 

[Mark44]

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.

 

[Peter G.]

Oh, ok, so when it is n N 3 for example, it should be n(n-1)(n-2)?

 

[Mark44]

Do you mean nC3? That would be n!/[3! (n - 3)!] = [n(n - 1)(n - 2)]/3!

 

[Peter G.]

Yeah, sorry, that was what I meant. Thanks!

 

[Mark44]

This seems to be harder than it should be - are you sure you have the right coefficients?

 

[Peter G.]

Yes, the answers the book has are: n = 16 and a = -0.5

 

[Mark44]

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.