Step 1: Express the product using double-angle formulas.
We want to expand $\sin^4\theta \cos^2\theta$ in terms of cosines of multiples of $\theta$. Write $\sin^4\theta \cos^2\theta = \sin^2\theta \cdot \sin^2\theta \cos^2\theta$.
Step 2: Use the identities $\sin^2\theta = \frac{1-\cos 2\theta}{2}$ and $\sin^2\theta\cos^2\theta = \frac{\sin^2 2\theta}{4}$.
\[\sin^4\theta\cos^2\theta = \frac{1-\cos 2\theta}{2} \cdot \frac{\sin^2 2\theta}{4} = \frac{(1-\cos 2\theta)\sin^2 2\theta}{8}.\]
Step 3: Expand using $\sin^2 2\theta = \frac{1-\cos 4\theta}{2}$.
\[\sin^4\theta\cos^2\theta = \frac{1}{8}\left(\frac{1-\cos 4\theta}{2} - \cos 2\theta \cdot \frac{1-\cos 4\theta}{2}\right) = \frac{1}{16}\bigl[(1-\cos 4\theta) - \cos 2\theta(1-\cos 4\theta)\bigr].\]
Step 4: Further simplify using product-to-sum.
$\cos 2\theta \cos 4\theta = \frac{1}{2}[\cos 6\theta + \cos 2\theta]$. So: \[\sin^4\theta\cos^2\theta = \frac{1}{16}\left[1 - \cos 4\theta - \cos 2\theta + \frac{\cos 6\theta + \cos 2\theta}{2}\right]\] \[= \frac{1}{16}\left[1 - \frac{\cos 2\theta}{2} - \cos 4\theta + \frac{\cos 6\theta}{2}\right].\]
Step 5: Identify the series $\sum a_{2n}\cos(2n\theta)$ and its coefficients.
Reading off: $a_0 = \frac{1}{16}$, $a_2 = -\frac{1}{32}$, $a_4 = -\frac{1}{16}$, $a_6 = \frac{1}{32}$, and $a_{2n} = 0$ for all $n \geq 4$.
Step 6: Find the least $n$ for which $a_{2n} = 0$.
The first $n$ (with $n \geq 1$) for which $a_{2n} = 0$ is $n = 4$, since $a_2, a_4, a_6$ are all nonzero but $a_8 = 0$. \[ \boxed{n = 4} \]