Finding binomial Coefficient using expansion If ${\left( {1 + x + {x^2}} \right)^n} = {a_0} + {a_1}x + {a_2}{x^2} + ... + {a_{2n - 1}}{x^{2n - 1}} + {a_{2n}}{x^{2n}}$, prove that $a_0=a_{2n},a_1=a_{2n-1},....a_n=a_{n+1}$
My approach is as follow ${\left( {1 + x + {x^2}} \right)^n} = {a_0} + {a_1}x + {a_2}{x^2} + ... + {a_{2n - 1}}{x^{2n - 1}} + {a_{2n}}{x^{2n}}$
${\left( {1 + x + {x^2}} \right)^n} = {\left( {1 + X} \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{X^r}}  = \sum\limits_{r = 0}^n {{}^n{C_r}{{\left( {x + {x^2}} \right)}^r}}  = \sum\limits_{r = 0}^n {{}^n{C_r}\sum\limits_{g = 0}^r {{}^r{C_g}{x^{r - g}}.{x^{2g}}} }  = \sum\limits_{r = 0}^n {{}^n{C_r}\sum\limits_{g = 0}^r {{}^r{C_g}{x^{r + g}}} } $
How do we proceed from here

$$(1+x+x^2)^n=\sum_{k=0}^{2n} A_k x^k$$                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  Change $x$ to $1/x$ in this udentity, then
$$(1+x+x^2)^n=\sum_{k=0}^{2n} A_k x^{2n-k}=\sum_{k=0}^{2n} A_{2n-k} ~ x^k,$$ $k \to 2n-k$ used in above. Finally we get $A_{2n-k}=A_k$.
