For $(1+x+x^2)^n = A_0 + A_1x + ... + A_{2n}x^{2n}$, prove that $(n-r)A_r + (2n -r+1)A_{r-1} = (r+1)A_{r+1}$

Question

My try: One way to do this:

Differentiate the original expression

Divide the resultant expression with the original expression

Compare coefficients of $A_r$ on both sides

This will give the result.

Is there any other way(a more elegant one perhaps) to derive this result?

Answer 1

Hint. By induction: compute $$(n+1-r)A^{n+1}_r+(2n+2-r+1)A^{n+1}_{r-1}$$ and write it as $(r+1)A^{n+1}_{r+1}$ using the relation $A^{n+1}_r=A^n_r+A^n_{r-1}+A^n_{r-2}$. If you group the terms smartly, this takes only a few lines...