(0)

\[ S_{j}\left(n\right)=\sum_{k=0}^{\infty}C\left(3n,3k+j\right) \] とおく。
そうすると、
\begin{align*} S_{0}\left(n\right) & =\sum_{k=0}^{\infty}C\left(3n,3k\right)\\ & =\sum_{k=0}^{\infty}\left\{ C\left(3n-3,3k\right)+3C\left(3n-3,3k-1\right)+3C\left(3n-3,3k-2\right)+C\left(3n-3,3k-3\right)\right\} \\ & =S_{0}\left(n-1\right)+3S_{2}\left(n-1\right)+3S_{1}\left(n-1\right)+S_{0}\left(n-1\right)\\ & =3\left\{ S_{0}\left(n-1\right)+S_{2}\left(n-1\right)+S_{1}\left(n-1\right)\right\} -S_{0}\left(n-1\right)\\ & =3\sum_{k=0}^{\infty}C\left(3\left(n-1\right),k\right)-S_{0}\left(n-1\right)\\ & =3\cdot2^{3\left(n-1\right)}-S_{0}\left(n-1\right)\\ & =\left(-1\right)^{n}\sum_{k=1}^{n}\left\{ \left(-1\right)^{k}S_{0}\left(k\right)-\left(-1\right)^{k-1}S_{0}\left(k-1\right)\right\} +\left(-1\right)^{n}S_{0}\left(0\right)\\ & =\left(-1\right)^{n}\sum_{k=1}^{n}\left(-1\right)^{k}3\cdot2^{3\left(k-1\right)}+\left(-1\right)^{n}\\ & =\left(-1\right)^{n}\frac{3}{2^{3}}\sum_{k=1}^{n}\left(-2^{3}\right)^{k}+\left(-1\right)^{n}\\ & =\left(-1\right)^{n+1}3\frac{1-\left(-2^{3}\right)^{n}}{1+2^{3}}+\left(-1\right)^{n}\\ & =\frac{2^{3n}-\left(-1\right)^{n}}{3}+\left(-1\right)^{n}\\ & =\frac{2^{3n}+2\left(-1\right)^{n}}{3} \end{align*} となるので、
\[ \sum_{k=0}^{\infty}C\left(3n,3k\right)=\frac{2^{3n}+2\left(-1\right)^{n}}{3} \] となる。

(0)-2

\begin{align*} \sum_{k=0}^{3n}C\left(3n,k\right)\omega^{k} & =\left(1+\omega\right)^{3n}\\ & =\left(-\omega^{2}\right)^{3n}\\ & =\left(-1\right)^{n} \end{align*} となり、別の計算をすると、
\begin{align*} \sum_{k=0}^{3n}C\left(3n,k\right)\omega^{k} & =\sum_{k=0}^{n}C\left(3n,3k\right)\omega^{3k}+\sum_{k=0}^{n-1}\left\{ C\left(3n,3k+1\right)\omega^{3k+1}+C\left(3n,3k+2\right)\omega^{3k+2}\right\} \\ & =\sum_{k=0}^{n}C\left(3n,3k\right)+\sum_{k=0}^{n-1}\left\{ C\left(3n,3k+1\right)\omega+C\left(3n,3n-\left(3k+1\right)\right)\omega^{2}\right\} \\ & =\sum_{k=0}^{n}C\left(3n,3k\right)+\sum_{k=0}^{n-1}\left\{ C\left(3n,3k+1\right)\left(\omega+\omega^{2}\right)\right\} \\ & =\sum_{k=0}^{n}C\left(3n,3k\right)-\sum_{k=0}^{n-1}C\left(3n,3k+1\right) \end{align*} となる。
これらより、
\begin{align*} \sum_{k=0}^{n-1}C\left(3n,3k+1\right) & =\sum_{k=0}^{n}C\left(3n,3k\right)-\sum_{k=0}^{3n}C\left(3n,k\right)\omega^{k}\\ & =\sum_{k=0}^{n}C\left(3n,3k\right)-\left(-1\right)^{n} \end{align*} となる。
これを使うと、
\begin{align*} \sum_{k=0}^{n}C\left(3n,3k\right) & =\sum_{k=0}^{3n}C\left(3n,k\right)-\sum_{k=0}^{n-1}\left\{ C\left(3n,3k+1\right)+C\left(3n,3k+2\right)\right\} \\ & =\sum_{k=0}^{3n}C\left(3n,k\right)-\sum_{k=0}^{n-1}\left\{ C\left(3n,3k+1\right)+C\left(3n,3n-\left(3k+1\right)\right)\right\} \\ & =\sum_{k=0}^{3n}C\left(3n,k\right)-2\sum_{k=0}^{n-1}C\left(3n,3k+1\right)\\ & =2^{3n}-2\left(\sum_{k=0}^{n}C\left(3n,3k\right)-\left(-1\right)^{n}\right)\\ & =\frac{2^{3n}+2\left(-1\right)^{n}}{3} \end{align*} となる。