Binom

We know that $\displaystyle (1+x)^n = \sum_{k=0}^n {n \choose k}x^k$ . Let $\omega$ be the third root of unit. Then we have $1+\omega+\omega^2=0$ and $\omega^3 = 1$ .

$\begin{aligned} (1+1)^{2010} & = {2010 \choose 0} + {2010 \choose 1} + {2010 \choose 2} + {2010 \choose 3} + \cdots + {2010 \choose 2010} \\ (1+\omega)^{2010} & = {2010 \choose 0} + {2010 \choose 1}\omega + {2010 \choose 2}\omega^2 + {2010 \choose 3} + \cdots + {2010 \choose 2010} \\ (1+\omega^2)^{2010} & = {2010 \choose 0} + {2010 \choose 1}\omega^2 + {2010 \choose 2}\omega + {2010 \choose 3} + \cdots + {2010 \choose 2010} \end{aligned}$

Therefore,

$\begin{aligned} (1+1)^{2010} + (1+\omega)^{2010} + (1+\omega^2)^{2010} & = 3 \left( {2010 \choose 0} + {2010 \choose 3} + {2010 \choose 6} + \cdots + {2010 \choose 2010} \right) \end{aligned}$

$\begin{aligned} \implies \sum_{k=0}^{670} {2010 \choose 3k} & = \frac {(1+1)^{2010} + (1+\omega)^{2010} + (1+\omega^2)^{2010}}3 & \small \color{#3D99F6} \text{Since } 1+\omega+\omega^2=0 \\ & = \frac {2^{2010} + (-\omega^2)^{2010} + (-\omega)^{2010}}3 \\ & = \frac {2^{2010} + \omega^{4020} + \omega^{2010}}3 \\ & = \frac {2^{2010} + 1 + 1}3 \\ & = \frac {2^{2010} + 2}3 \end{aligned}$

$\implies n = \boxed{2010}$