Binomial series Question

This might have been asked before, but I'm stuck on this one for a while now:

(1+px+qx²)^-2 has coefficients of 4 and 14 for x and x² respectively.

Find p and q.

Thanks for any solutions :smile:

Reply 1

john !!

1/(1+px+qx²)

hmm.. i havent done binomial with a negative index. perhaps you could factorise and split into partial fractions?

put:
qx² + px + 1 = 0
x² + (p/q)x = -1/q
[x + (p/2q)]² = (p²-4q)/4q²
x + (p/2q) = ±[√(p²-4q)]/2q
x = [-p±√(p²-4q)]/2q

so

1/(1+px+qx²)² = 1/[(x-[-p+√(p²-4q)]/2q)(x-[-p-√(p²-4q)]/2q)]²
= 1/[(x+[p-√(p²-4q)]/2q)(x+[p+√(p²-4q)]/2q)]²
= 1/{x+[p-√(p²-4q)]/2q}²{x+[p+√(p²-4q)]/2q}²

= A/{x+[p-√(p²-4q)]/2q} + B/{x+[p-√(p²-4q)]/2q}² + C/{x+[p+√(p²-4q)]/2q} + D/{x+[p+√(p²-4q)]/2q}²

hmm maybe not.

Reply 2

m:)ckel

I'm not sure about this one, because the 3 terms is bugging me slightly. I get:
p = 2 , q = -1
But, I'm not certain on that at all. Do you have the answers?

Reply 3

Freddy C

Well, thanks for the attempt to both.

The answer is P = -2 and q = -1

I got P easily =>

the second term is px (or nx as it's normally called in textbooks) and since the coeff. is 4 and so

-2p = 4

p = -2

were you meant to type "p = 2" or did you actually mean "p = -2" mockel?

q is -1, how did you get that?

Reply 4

m:)ckel

I actually meant to type p=2, but have seen a slight error in my working, and so I get p=-2.
I'll just show you the whole thing (since I've already typed it out somewhere else, but for the wrong question!)

What I did was just to expand normally, but keeping the 'x' and 'x^2' terms together (if you know what I mean)

(1+px+qx^2)^-2
= 1 + (-2)(px+qx^2) + [(-2)(-3)/2](px+qx^2)^2 + ...

This is all that's needed since there are no more x^2 terms.

Expanding, you get:

1 - 2px + (3p^2 - 2q)x^2 + ...

Equating coefficients:

-2p = 4
p = -2

3p^2 - 2q = 14
12 - 2q = 14
q = -1

Reply 5

Freddy C