Let's calculate the sum

$$
\sum_{j=0}^{n}\binom{2 n}{2 j}(-3)^{j}
$$
# I. Solution. Let

$$
S_{m}=\sum_{j=0}^{[m / 2]}\binom{m}{2 j}(-3)^{j}
$$

We claim that the value of $S_{2 n}$ is $2^{2 n}$ if $n$ is divisible by 3, and $-2^{2 n-1}$ otherwise. We prove this by complete induction on $n$. The cases $n=1,2,3$ can be easily verified to hold:

$$
S_{2}=1-3 \cdot 1=-2, \quad S_{4}=1-3 \cdot 6+9 \cdot 1=-8, \quad S_{6}=1-3 \cdot 15+9 \cdot 15-27 \cdot 1=64
$$

For the induction, it suffices to show that for $m>3$, $S_{m}=(-8) S_{m-3}$, since this implies $S_{2 n}=(-8) S_{2 n-3}=64 S_{2(n-3)}$, from which the claim follows immediately. To prove $S_{m}=(-8) S_{m-3}$, we first express the binomial coefficient $\binom{m}{2 k}$ as a sum of binomial coefficients from the $(m-3)$-rd row of Pascal's triangle:

$$
\begin{aligned}
\binom{m}{2 j}= & \binom{m-1}{2 j-1}+\binom{m-1}{2 j}=\binom{m-2}{2 j-2}+2\binom{m-2}{2 j-1}+\binom{m-2}{2 j}= \\
& =\binom{m-3}{2 j-3}+3\binom{m-3}{2 j-2}+3\binom{m-3}{2 j-1}+\binom{m-3}{2 j}
\end{aligned}
$$

where we consider the value of $\binom{a}{b}$ to be 0 if $b<0$ or $a<b$. Using this,

$$
S_{m}=\sum_{j=0}^{[m / 2]}\left(\binom{m-3}{2 j-3}+3\binom{m-3}{2 j-2}+3\binom{m-3}{2 j-1}+\binom{m-3}{2 j}\right)(-3)^{j}
$$

We calculate the coefficient of $\binom{m-3}{\ell}$ in this sum for $0 \leq \ell \leq m-3$. If $\ell=2 k+1$ is an odd number, the coefficient is

$$
1 \cdot(-3)^{k+2}+3 \cdot(-3)^{k+1}=0
$$

If $\ell=2 k$ is even, the coefficient is

$$
3 \cdot(-3)^{k+1}+1 \cdot(-3)^{k}=-8 \cdot(-3)^{k}
$$

Thus,

$$
S_{m}=\sum_{k=0}^{[(m-3) / 2]}\binom{m-3}{2 k}(-8)(-3)^{k}=-8 S_{m-3}
$$

This proves that the value of the sum is $2^{2 n}$ if $n$ is divisible by 3, and $-2^{2 n-1}$ otherwise.

II. Solution. Consider the complex number

$$
z=1+\sqrt{3} i=2\left(\cos \frac{\pi}{3}+i \cdot \sin \frac{\pi}{3}\right)
$$

and determine its $2 n$-th power (where $n$ is a positive integer). Using the trigonometric form and De Moivre's formula, we get

$$
z^{2 n}=2^{2 n}\left(\cos \left(n \cdot \frac{2 \pi}{3}\right)+i \cdot \sin \left(n \cdot \frac{2 \pi}{3}\right)\right)
$$

On the other hand, using the algebraic form and the binomial theorem:

$$
z^{2 n}=\sum_{j=0}^{2 n}\binom{2 n}{j} \cdot 1^{2 n-j} \cdot(\sqrt{3} i)^{j}=\sum_{j=0}^{n}\binom{2 n}{2 j}(-3)^{j}+i \sqrt{3} \sum_{j=0}^{n-1}\binom{2 n}{2 j+1}(-3)^{j}
$$

By comparing the real parts of the two expressions for $z^{2 n}$,

$$
\sum_{j=0}^{n}\binom{2 n}{2 j}(-3)^{j}=2^{2 n} \cdot \cos \left(n \cdot \frac{2 \pi}{3}\right)
$$

follows.