A finite series

Let $N:=2021$ and define the sequence $a_k:=(-1)^{k+1}\binom{N}{k}$ . If we wanted to calculate the sum over the entire sequence $a_k$ , the task would be quite easy. However, we are only interested in every third term! What we need is a filter that takes in the entire sequence $a_k$ and sets all unwanted terms to zero. Luckily, there is a thing called

Unity-Filter: Let $n\in\mathbb{N}$ and define $z=e^{i\frac{2\pi}{n}}$ as the fundamental n'th root of unity. Then $k \in\mathbb{Z}:\qquad u_k^{(n)}:=\frac{1}{n}\sum_{l=0}^{n-1}z^{kl} = \begin{cases} 1:&k\in n\mathbb{Z}\\ 0:&\text{else} \end{cases}$

If we multiply the unity-filter with a sequence, it only lets every n'th element pass and sets all other elements to zero. In this task, we need a unity-filter with $n=3$ to rewrite the sum $X$ we need to calculate. Adding $a_0=-1$ , we get $\begin{aligned} X -1&:=\sum_{k=0}^{\lfloor\frac{N}{3}\rfloor}a_{3k}=\sum_{k=0}^Nu^{(3)}_ka_k=\frac{1}{3}\sum_{l=0}^2\sum_{k=0}^N\binom{N}{k}(-1)^{k+1}z^{lk}=-\frac{1}{3}\sum_{l=0}^2 (1-z^l)^N&\left| \text{Binomial Theorem} \right. \end{aligned}$ The term with $l=0$ vanishes, and the other two are complex conjugates because $z^2=z^*$ : $\begin{aligned} X&=1-\frac{2}{3}\Re\left\{(1-z)^N\right\}=1-\frac{2}{3}\cdot 3^{\frac{N}{2}}\cos\left(-N\frac{\pi}{6}\right)=1+ 3^{\frac{N-1}{2}} &\left| 1-z=1-e^{i\frac{2\pi}{3}}=\sqrt{3}e^{-i\frac{\pi}{6}} \right. \end{aligned}$ While it is possible to calculate $X$ with arbitrary size integers (e.g. with maxima), we can also use logarithms to find $a,\:k$ : $\begin{aligned} k&=\left\lfloor\log_{10}\left( 1 + 3^{\frac{N-1}{2}}\right)\right\rfloor=:\lfloor k'\rfloor = 481, &&& a&=\lceil 10^{k'-k}\rceil=8&&&\Rightarrow &&&& a\times k&=\boxed{3848} \end{aligned}$