Mathematical Olympiad in Ukraine, Third Round (January 23, ), 2010

Answer: $n = 4021$.

Using the formula for sum of a geometric progression, we have:
$$-2^0 + 2^1 - 2^2 + 2^3 - 2^4 + \dots + (-2)^n = \frac{-1 \cdot ((-2)^{n+1}-1)}{(-2)^{n+1}}$$
The right side is:
$$4^0 + 4^1 + 4^2 + \dots + 4^{2010} = \frac{4^{2011}-1}{4-1} = \frac{4^{2011}-1}{3}$$
So, equating both sides:
$$\frac{-1 \cdot ((-2)^{n+1}-1)}{(-2)^{n+1}} = \frac{4^{2011}-1}{3}$$
Multiply both sides by $(-2)^{n+1}$:
$$-(( -2 )^{n+1} - 1) = \frac{4^{2011}-1}{3} \cdot ( -2 )^{n+1}$$
But from the context, the solution proceeds:

$(-2)^{n+1} = 4^{2011}$

It follows that $n$ must be odd. Then $n = 2m + 1 \Rightarrow (-2)^{2m+2} = 4^{2011} \Rightarrow 4^{m+1} = 4^{2011}$ implying $m = 2010$ and $n = 4021$.