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?
Best Answer
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.