If $\sum_{n = 0}^{\infty}\cos^{2n}\theta = 5$, what is the value of $\cos{2\theta}$?
From the formula for an infinite geometric series,
\[\sum_{n = 0}^\infty \cos^{2n} \theta = 1 + \cos^2 \theta + \cos^4 \theta + \dotsb = \frac{1}{1 - \cos^2 \theta} = 5.\]Hence, $\cos^2 \theta = \frac{4}{5}.$  Then
\[\cos 2 \theta = 2 \cos^2 \theta - 1 = \boxed{\frac{3}{5}}.\]
\frac{3}{5}
Level 4
Precalculus
Problem: If $\sum_{n = 0}^{\infty}\cos^{2n}\theta = 5$, what is the value of $\cos{2\theta}$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the problem, we start with the infinite series given by $\sum_{n = 0}^{\infty}\cos^{2n}\theta$. This series is a geometric series where the first term is 1 (when n=0) and the common ratio is $\cos^2\theta$. The formula for the sum of an infinite geometric series is $\frac{a}{1 - r}$, where $a$ is the first term and $r$ is the common ratio. 

Here, we have:
- First term $a = 1$
- Common ratio $r = \cos^2\theta$

Thus, the sum can be expressed as:
$$\sum_{n = 0}^{\infty}\cos^{2n}\theta = \frac{1}{1 - \cos^2\theta}$$

We know from trigonometric identities that $1 - \cos^2\theta = \sin^2\theta$. Therefore, we can rewrite the sum as:
$$\sum_{n = 0}^{\infty}\cos^{2n}\theta = \frac{1}{\sin^2\theta}$$

According to the problem, this sum equals 5:
$$\frac{1}{\sin^2\theta} = 5$$

To find $\sin^2\theta$, we take the reciprocal:
$$\sin^2\theta = \frac{1}{5}$$

Next, we use the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ to find $\cos^2\theta$:
$$\cos^2\theta = 1 - \sin^2\theta = 1 - \frac{1}{5} = \frac{4}{5}$$

Now, we need to find $\cos{2\theta}$. The double angle formula for cosine is:
$$\cos{2\theta} = 2\cos^2\theta - 1$$

Substituting the value of $\cos^2\theta$ we found:
$$\cos{2\theta} = 2\left(\frac{4}{5}\right) - 1 = \frac{8}{5} - 1 = \frac{8}{5} - \frac{5}{5} = \frac{3}{5}$$

Thus, the value of $\cos{2\theta}$ is $\frac{3}{5}$.<|END_OF_LATENT|> The final answer is $\frac{3}{5}$.