I presumed that $a_n$ in the second part is a typo error hence I modified the question
The question is as follow
If ${\left( {1 + x + {x^2}} \right)^n} = {a_0} + {a_1}x + {a_2}{x^2} + .. + {a_{2n - 1}}{x^{2n - 1}} + {a_{2n}}{x^{2n}}$ then ${\left( {{a_0} + {a_1} + {a_2} + ... + {a_n}} \right)^3} - {\left( {{a_{n + 1}} + .. + {a_{2n - 1}} + {a_{2n}}} \right)^3}$ is (where $n\in N$)
(A) $0$
(B) $\frac{3^n-1}{2}$
(C) $\frac{9^n+1}{2}$
(D) $\frac{9^n-1}{2}$
My approach is as follow
Put x=1, ${3^n} = {a_0} + {a_1} + {a_2} + .. + {a_{2n - 1}} + {a_{2n}} = T + U$
$T = \left( {{a_0} + {a_1} + {a_2} + ... + {a_n}} \right)\& U = \left( {{a_{n + 1}} + .. + {a_{2n - 1}} + {a_{2n}}} \right)$
$G = {\left( {{a_0} + {a_1} + {a_2} + ... + {a_n}} \right)^3} - {\left( {{a_{n + 1}} + .. + {a_{2n - 1}} + {a_{2n}}} \right)^3} = {T^3} - {U^3}$
$G = \left( {T - U} \right)\left( {{T^2} + {U^2} + UT} \right) = \left( {T - U} \right)\left( {{{\left( {T + U} \right)}^2} - UT} \right) = \left( {T - U} \right)\left( {{3^{2n}} - UT} \right) = \left( {T - U} \right)\left( {{9^n} - UT} \right)$
${\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}.{x^r}{{\left( {1 + x} \right)}^r}} = \sum\limits_{r = 0}^n {{}^n{C_r}.{x^r}\sum\limits_{g = 0}^r {{}^r{C_g}.{x^g}} } $
The solution is given as "Equidistant terms from beginning and end have the same values. Hence $a_0=a_{2n}$ &$a_1=a_{2n-1}$ "
Hence T=U, but how do I prove from the above binomial theorem but getting the general term