Bounds for $S_k = \sum_{i=1}^k {3k \choose 3i}$ Consider:
$$S_k = \sum_{i=0}^k {3k \choose 3i}.$$

Is it true that for all sufficiently large values of $k$, $S_k(1/2)^{3k} < 1/2$?

In general, for: 
$$S_{c,k} {ck \choose ci} = \sum_{i=0}^k,$$
for integer $c > 1$.

Is it true that for all sufficiently large values of $k$, $S_{c,k}(1/2)^{ck} < 1/(c-1)$?


There actually is a closed form for this sum.
Let $\lambda \in \mathbb{C}$. By the binomial formula we know that
$$\sum_{i=0}^{3k} \binom{3k}{i} \cdot \lambda^i = (1+\lambda)^{3k}.$$
We can view this in terms of linear algebra: let
$$\begin{align*}
u & = \left[ \binom{3k}{0}, \binom{3k}{1}, \binom{3k}{2}, \ldots, \binom{3k}{3k} \right] \\[1ex]
v & = [1, \lambda, \lambda^2, \ldots, \lambda^{3k}]
\end{align*}$$
so that $u, v \in \mathbb{R}^{3k+1}$ and $\left< u, v \right> = (1+\lambda)^{3k}$. Now let $\omega = \cos \frac{2 \pi}{3} + i \sin \frac{2 \pi}{3}$ be the $3$rd root of unity and substitute $\alpha = 1, \ \omega, \ \omega^2$ in the definition of $v$ to get $v_0, v_1, v_2$ respectively. Now consider 
$$w = [1, 0, 0, 1, 0, 0, 1, \ldots, 1] \in \mathbb{R}^{3k+1}.$$
Then our sum can expressed as 
$$\sum_{i=0}^k \binom{3k}{3i} = \left< u, w \right>$$
but it's easy to see that $w \in \operatorname{span} \{ v_0, v_1, v_2 \}$, hence there are $\alpha_0, \alpha_1, \alpha_2 \in \mathbb{C}$ such that 
$$w = \alpha_0 v_0 + \alpha_1 v_1 + \alpha_2 v_2.$$
Then
$$\left< u, w \right> = \alpha_0 \left< u, v_0 \right> + \alpha_1 \left< u, v_1 \right> + \alpha_2 \left< u, v_2 \right> = \alpha_0 2^{3k} + \alpha_1 (1+\omega)^{3k} + \alpha_2 (1+\omega^2)^{3k}.$$
This easily generalizes to finding the sum $\displaystyle \sum_{i=0}^k \binom{ck}{ci}$ for any positive integer $c$, namely
$$\sum_{i=0}^k \binom{ck}{ci} = \sum_{j=0}^{c-1} \alpha_j \cdot (1+\omega^j)^{3k}$$
where $\omega = \cos \frac{2 \pi}{c} + i \sin \frac{2 \pi}{c}$ and the coeffiecients $\alpha_j$ are found by solving an analogous system of equations.
I don't know if this helps finding real lower bounds on the sum though.
