To find the sum: $\frac {1}{n!} \sum \binom {n}{2+3r} x^{1+r}$ 
Sum the series:
$$
\frac {x}{2!(n-2)!}+\frac {x^2}{5!(n-5)!}+\frac {x^3}{8!(n-8)!}+....+\frac {x^{\frac {n}{3}}}{(n-1)!},
$$
$n$ being a multiple of $3$.(Math. Tripos, 1899)

My attempt
We may write for an integer $m$, $n=3m$ so that:
$$
S(x)=\frac {x}{2!(n-2)!}+\frac {x^2}{5!(n-5)!}+\frac {x^3}{8!(n-8)!}+....+\frac {x^{\frac {n}{3}}}{(n-1)!}
$$
$$
=\frac {x}{2!(3m-2)!}+\frac {x^2}{5!(3m-5)!}+\frac {x^3}{8!(3m-8)!}+....+\frac {x^m}{(3m-1)!}
$$
$$
=\frac {1}{(3m)!}\left\{ \binom {3m}{2}x+\binom {3m}{5}x^2+\binom {3m}{8}x^3+....+\binom {3m}{3m-1}x^m \right\}
$$
Here it is plain to me that it is some binomial expansion in which every $2$ out of $3$ terms cancel out. The first guess in such a situation is a series consisting the cube roots of unity as well as $\sqrt[3]x.$
$$
(\omega_3+\sqrt[3]x)^{3m}=1+\binom {3m}{1}\omega_3^2\sqrt[3]x+\binom {3m}{2}\omega_3\sqrt[3]{x^2}+....+x^m
$$
$$
S(x)=\frac {\omega_3^2 \sqrt[3]x}{(3m)!}\left\{ \binom {3m}{2}\omega_3\sqrt[3]{x^2}+\binom {3m}{5}\omega_3x^2\sqrt[3]{x^2}+....+\binom {3m}{3m-1}\omega_3x^{m-1}\sqrt[3]{x^2}\right\}
$$
Now if,
$$
S(x)=\frac {\omega_3^2 \sqrt[3]x}{(3m)!} (\omega_3+\sqrt[3]x)^{3m},
$$
then we are left with two conditions:
$$
\binom {3m}{1}+\binom {3m}{4}x+\binom {3m}{7}x^2+....=0,
$$
$$
1+\binom {3m}{3}x+\binom {3m}{6}x^2+....=0
$$
It seems that the whole line of logic is circular. Any help would be greatly appreciated.

Hint
Consider
$$S=a(1+y)^n+b(1+wy)^n+(1+w^2y)^n$$ where $w$ is a complex cube (as we need every third term) root of unity
We need the coefficients of $y^0,y^1$ to be zero
$a+b+1=0$
$an+bnw+w^2n=0$
Solve for $a,b$
Now find $S$ and compare with given expression
