# Binomial expansion

1. Nov 10, 2014

### erisedk

1. The problem statement, all variables and given/known data
Consider the expansion (ax2 + bx + c)n = ∑(r=0 to r=2n) Ar xr------------------(1) , where Ar is real ∀ 0 ≤ r ≤ 2n
Replacing x by c/(ax) and using the property ∑(r=0 to r=2n) Tr = ∑(r=0 to r=2n) T2n-r ,
we get (ax2 + bx + c)n = ∑(r=0 to r=2n) Br xr ----------------------(2).
Hence we get a relation between Ar and Br by comparing like power of xr
0 ≤ r ≤ 2n from both equations.

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

2. Relevant equations

3. 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.

2. Nov 10, 2014

### 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.

3. Nov 10, 2014

### 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.