回復 13# mcgrady0628 的帖子
第11題
若(1-x+x^2)^{1000}的展開式為a_0+a_1x+a_2x^2+\ldots+a_{2000}x^{2000},則a_0+a_3+a_6+a_9+\ldots+a_{1998}之值為 。
[解答]
\begin{align}
& {{a}_{0}}+{{a}_{1}}+{{a}_{2}}+\cdots +{{a}_{1998}}+{{a}_{1999}}+{{a}_{2000}}={{\left( 1+1+1 \right)}^{1000}}={{3}^{1000}} \\
& {{a}_{0}}+{{a}_{1}}\omega +{{a}_{2}}{{\omega }^{2}}+\cdots +{{a}_{1998}}+{{a}_{1999}}\omega +{{a}_{2000}}{{\omega }^{2}}={{\left( 1+\omega +{{\omega }^{2}} \right)}^{1000}}=0 \\
& {{a}_{0}}+{{a}_{1}}{{\omega }^{2}}+{{a}_{2}}\omega +\cdots +{{a}_{1998}}+{{a}_{1999}}{{\omega }^{2}}+{{a}_{2000}}\omega ={{\left( 1+{{\omega }^{2}}+\omega \right)}^{1000}}=0 \\
& 3\left( {{a}_{0}}+{{a}_{3}}+{{a}_{6}}+\cdots +{{a}_{1998}} \right)={{3}^{1000}} \\
& {{a}_{0}}+{{a}_{3}}+{{a}_{6}}+\cdots +{{a}_{1998}}={{3}^{999}} \\
\end{align}