Find the values of $a$ and $b$ when the binomial expansion of $\frac{1}{(1+ax)^b}+\frac{1}{(1+bx)^a} = 2-6x+15x^2$

Question

Find the values of $a$ and $b$ when the binomial expansion of $$\frac{1}{(1+ax)^b}+\frac{1}{(1+bx)^a} = 2-6x+15x^2$$ So I set up the two equations: $$(-b)(ax)+(-a)(bx)=-6x$$ and $$\frac{(-b)(-b-1)(ax)^2}{2}+\frac{(-a)(-a-1)(bx)^2}{2}=15x^2$$

I then substituted the first one into the second one to try and solve for $a$, but this hasn't given me the right answer Am I doing something wrong?

Answer 1

You've done almost everything already:
We have that $$(1+x)^n=1+nx+\frac{1}{2}n(n-1)x^2+O(x^3)$$ So $$\frac{1}{(1+ax)^b}+\frac{1}{(1+bx)^a}=$$ $$(1+ax)^{-b}+(1+bx)^{-a}=$$ $$\left(1-bax+\frac{1}{2}b(b+1)a^2x^2+O(x^3)\right)+\left(1-abx+\frac{1}{2}a(a+1)b^2x^2+O(x^3)\right)=$$ $$2-2abx+\frac{1}{2}(a^2b^2+ba^2+a^2b^2+ab^2)x^2+O(x^3)$$ So we need to have that $$-2ab=-6$$ $$ab=3$$ And $$\frac{1}{2}(a^2b^2+ba^2+a^2b^2+ab^2)=15$$ $$2(ab)^2+ab(a+b)=30$$ $$2(3)^2+3(a+b)=30$$ $$a+b=4$$ And I think you can solve the system of equations now.

Answer 2

Hint:

Compute $ab$ and $a+b$, and use the following result from high school:

Let $x, y$ two (real or complex) numbers with given sum $s$ and product $p$. Then $x$ and $y$ are the roots of the quadratic equation: $$t^2-st+p=0.$$

Answer 3

Find the derivative at $x=0$: $$\frac{1}{(1+ax)^b}+\frac{1}{(1+bx)^a} = 2-6x+15x^2 \Rightarrow \\ -\frac{ab}{(1+ax)^{b+1}}-\frac{ab}{(1+bx)^{a+1}}=-6+30x \Rightarrow ab=3.$$ Find the 2nd order derivative at $x=0$: $$\frac{a^2b(b+1)}{(1+ax)^{b+2}}+\frac{ab^2(a+1)}{(1+bx)^{a+2}}=30 \Rightarrow a+b=4.$$ Now you can solve the equations: $ab=3$ and $a+b=4$.