Finding coefficient ratio A₆/A₁₄ in (2x² + 3x + 4)¹⁰

[erisedk]

Homework Statement

Consider the expansion (ax² + bx + c)ⁿ = ∑(r=0 to r=2n) A_r x^r------------------(1) , where A_r is real ∀ 0 ≤ r ≤ 2n
Replacing x by c/(ax) and using the property ∑(r=0 to r=2n) T_r = ∑(r=0 to r=2n) T_2n-r ,
we get (ax² + bx + c)ⁿ = ∑(r=0 to r=2n) B_r x^r ----------------------(2).
Hence we get a relation between A_r and B_r by comparing like power of x^r ∀
0 ≤ r ≤ 2n from both equations.

Let (2x² + 3x + 4)¹⁰ = ∑(r=0 to r=2n) A_r x^r then the value of
A₆ /A₁₄ is

Homework Equations

The Attempt at a Solution

I tried doing what the paragraph says--
I substituted (2/x) for x (c/ax = 2/x).
I got (8/x^2 + 6/x + 4)¹⁰
But I don't want a relation between A and B
I want the ratio of the coefficients of the terms with x^6 and x^14.
I tried doing it using binomial expansion by making cases for when the powers of (4+3x) and (2x^2) add up to 6 and 14.
For 6, I got 4 such cases. For 14, I got 8. However, solving all the Cs and stuff is very time consuming and doesn't seem the right way to go.

 

[RUber]

The important thing to recognize is that your $B_r$ should be equal to your $A_r$ since they are expanding the same polynomial.
By introducing the substitution, you can find a way to relate $B_{2n-r}$ to $A_r$ which works for r=6.

 

[RUber]

Keep it in the sum notation to see the relationships.
$\sum_{r=0}^{2n} A_r x^r$
$\sum_{r=0}^{2n} B_r \left(\frac{c}{ax}\right)^r=\sum_{r=0}^{2n} B_r \left(\frac{c}{ax}\right)^{2n-r}$
Do some algebra to equate the powers of r, and see what comes out.