Starting 2018 with a bang

By binomial theorem , we have $\displaystyle (1+x)^n = \sum_{k=0}^n \binom nr x^r$ . Since $\displaystyle \binom nr x^r = 0$ if $r > n$ , then $\displaystyle \sum_{k=0}^\infty \binom nr x^r = \sum_{k=0}^n \binom nr x^r$ . Now, let us consider when $x = 1, \omega, \omega^2$ , where $\omega$ is the complex cube root of unit.

$\begin{aligned} (1+1)^{2018} & = \binom {2018}0 \cdot 1 + \binom {2018}1 \cdot 1 + \binom {2018}2 \cdot 1 + \binom {2018}3 \cdot 1 + \binom {2018}4 \cdot 1 + \binom {2018}5 \cdot 1 + \binom {2018}6 \cdot 1 + \cdots + \binom {2018}{2018} \cdot 1 \\ (1+\omega)^{2018} & = \binom {2018}0 \cdot 1 + \binom {2018}1 \omega + \binom {2018}2 \omega^2 + \binom {2018}3 \cdot 1 + \binom {2018}4 \omega + \binom {2018}5 \omega^2 + \binom {2018}6 \cdot 1 + \cdots + \binom {2018}{2018} \omega^2 & \small \color{#3D99F6} \text{Note that }\omega^3 = 1 \\ (1+\omega^2)^{2018} & = \binom {2018}0 \cdot 1 + \binom {2018}1 \omega^2 + \binom {2018}2 \omega + \binom {2018}3 \cdot 1 + \binom {2018}4 \omega^2 + \binom {2018}5 \omega + \binom {2018}6 \cdot 1 + \cdots + \binom {2018}{2018} \omega \end{aligned}$

$\begin{aligned} \implies (1+1)^{2018} + (1+\omega)^{2018} + (1+\omega^2)^{2018} & = 3 \binom {2018}0 + 3\binom {2018}3 + 3\binom {2018}6 + 3 \binom {2018}9 + \cdots + 3 \binom {2018}{2016} & \small \color{#3D99F6} \text{Note that }1+ \omega + \omega^2 = 0 \end{aligned}$

$\begin{aligned} \implies \sum_{k=0}^n \binom {2018}{3k} & = \frac {(1+1)^{2018} + (1+\omega)^{2018} + (1+\omega^2)^{2018}}3 & \small \color{#3D99F6} \text{Note that }1+ \omega + \omega^2 = 0 \\ & = \frac {2^{2018} + (-\omega^2)^{2018} + (-\omega)^{2018}}3 & \small \color{#3D99F6} \text{Note that }\omega^3 = 1 \\ & = \frac {2^{2018} + \omega + \omega^2}3 \\ & = \frac {2^{2018} -1}3 \end{aligned}$

Therefore, $a+b+c+d = 2+2018-1+3 = \boxed{2022}$ .

We will attempt to use the concept of Fourier transforms in this problem.

First off, we will use the fact that

$\large\displaystyle\sum_{k=0}^{\infty} \binom{n}{k}x^k = (x+1)^n$

Since we want to only consider every 3rd element of this general summation, we look for ways to generate such summations which will cancel the other elements. So we will look for values of $x$ which will eliminate the unwanted elements.

If we take a look at the "weights" of the elements of the sequence which are considered, it looks like

$(1, 0, 0, 1, 0, 0 ... )$

Which means that this sequence repeats periodically every 3 elements. This is the weight sequence that we desire.

On the general equation, since all elements are considered, the weight sequence is

$(1, 1, 1, 1, 1... )$

Now, what are sequences which are powers of a single number which are of period 3? The simplest answer is the solution set to $x^3 = 1$ , one of which is the one shown above.

The other two are as follows:

$( 1, e^{\frac{2i \pi}{3}}, e^{-\frac{2i \pi}{3}} ...)$

$( 1, e^{-\frac{2i \pi}{3}}, e^{\frac{2i \pi}{3}} ...)$

Now we can construct the weight sequence we want!

$3 (1,0,0,1...) = (1,1,1...) + ( 1, e^{\frac{2i \pi}{3}}, e^{-\frac{2i \pi}{3}} ...) + (1, e^{-\frac{2i \pi}{3}}, e^{\frac{2i \pi}{3}} ...)$

(See what I did there?)

Translating these back to the summations, we get

$3 \large\displaystyle\sum_{k=0}^{\infty} \binom{n}{3k} = (1+1)^n + (1 + e^{\frac{2i \pi}{3}})^n + ( 1 +e^{-\frac{2i \pi}{3}})^n$

$3N = 2^n + (1 + e^{\frac{2i \pi}{3}})^n + ( 1 +e^{-\frac{2i \pi}{3}})^n$

Now, we express

$e^{\frac{2i \pi}{3}} = - \frac{1}{2} + i \frac{\sqrt{3}}{2}$

and

$e^{-\frac{2i \pi}{3}} = - \frac{1}{2} - i \frac{\sqrt{3}}{2}$

This simplifies it down to

$3N = 2^n + (1 - \frac{1}{2} + i \frac{\sqrt{3}}{2})^n + ( 1 + - \frac{1}{2} - i \frac{\sqrt{3}}{2} )^n$

$3N = 2^n + ( \frac{1}{2} + i \frac{\sqrt{3}}{2})^n + ( \frac{1}{2} - i \frac{\sqrt{3}}{2} )^n$

$3N = 2^n + (- e^{-\frac{2i \pi}{3}})^n + ( -e^{\frac{2i \pi}{3}})^n$

$3N = 2^n + (-1)^n [ e^{ \frac{2 i \pi n}{3}} + e^{\frac{-2i \pi n}{3}} ]$

$3N = 2^n + (-1)^n (2 \cos{ \frac{2\pi n}{3}})$

$3N = 2^n + 2(-1)^n ( \cos{ \frac{2\pi n}{3}})$

$N = \frac {2}{3} [ 2^{n-1} + (-1)^{n}( \cos{ \frac{2\pi n }{3}})]$

Plugging in $n =2018$ gives us

$N = \frac {2 }{3} [ 2^{2017} + (-1)^{2018}( \cos{ \frac{4036\pi }{3}})]$

$\large\boxed{N = \frac{2^{2018} -1}{3}}$