Combinatorics?

Starting with Binomial expansion of $(1+x)^n$

$\begin{aligned} (1+x)^{n} & = \dbinom{n}{0}+\dbinom{n}{1}x^1+\cdots+\dbinom{n}{n}x^n \\ n(1+x)^{n-1} & = 1\dbinom{n}{1}+2\dbinom{n}{2}x^{1}+\cdots+n\dbinom{n}{n}x^{n-1} & \small\color{#3D99F6}{\text{1st derivative}} \\ n\cdot 2^{n-1} & = \dbinom{n}{1}+2\dbinom{n}{2}+\cdots+n\dbinom{n}{n}& \small\color{#3D99F6}{\text{put x=1 }} \\ n(1+\omega)^{n} & = \dbinom{n}{1}+2\dbinom{n}{2}\omega^{2}+\cdots+n\dbinom{n}{n}\omega^{n-1} & \small\color{#3D99F6}{\text{put x= }} \omega \\ n(1+\omega^{2})^{n} & = \dbinom{n}{1}+2\dbinom{n}{2}\omega^{4}+\cdots+n\dbinom{n}{n}\omega^{2n-2} & \small\color{#3D99F6}{\text{put x= }} \omega^{2} \end{aligned}$

Add all the three equations

$\begin{aligned} n\left( 2^{n-1}+(1+\omega)^{n}+(1+\omega^{2})^{n} \right) & = 3\left[ 1\dbinom{n}{1}+4\dbinom{n}{4}+7\dbinom{n}{7}+\cdots+ \right] \\ \frac{n}{3}\left[2^{n-1}+2\cos{\left(\dfrac{(n-1)\pi}{3}\right)}\right] & = 1\dbinom{n}{1}+4\dbinom{n}{4}+7\dbinom{n}{7}+\cdots \\ \frac{97}{3}\left[2^{96}+2\cos{\left(\dfrac{96\pi}{3}\right)}\right] & = \displaystyle\sum_{r=1}^{33}\left(3r-2\right)\dbinom{97}{3r-2} \\ \frac{97}{3}\left[2^{96}+2\cos{\left(\dfrac{96\pi}{3}\right)}\right] & = \displaystyle\sum_{r=1}^{33}\left(3r-2\right)\dbinom{97}{3r-2} \\ \frac{97}{3}\left[ 2^{96}+2 \right] & = \displaystyle\sum_{r=1}^{33}\left(3r-2\right)\dbinom{97}{3r-2} \\ 4\times\frac{97}{9}\left[ 2^{94}+\dfrac{1}{2} \right] & = \displaystyle\sum_{r=1}^{33}\left(r-\frac{2}{3}\right)\dbinom{97}{3r-2} \end{aligned} \\ k = \boxed{\dfrac{388}{9}} \\ \lfloor k \rfloor = 43$