Let $(1+x+x^2)^n=a_1x+a_2x^2+ \cdots + a_{2n}x^{2n}$ be an identity in $x$. If we let $s=a_0+a_2+a_4+\cdots +a_{2n}$, then $s$ equals:
$\text{(A) } 2^n \quad \text{(B) } 2^n+1 \quad \text{(C) } \frac{3^n-1}{2} \quad \text{(D) } \frac{3^n}{2} \quad \text{(E) } \frac{3^n+1}{2}$
Let $f(x)=(1+x+x^2)^n$ then we have 
\[f(1)=a_0+a_1+a_2+...+a_{2n}=(1+1+1)^n=3^n\]
\[f(-1)=a_0-a_1+a_2-...+a_{2n}=(1-1+1)^n=1\]
Adding yields
\[f(1)+f(-1)=2(a_0+a_2+a_4+...+a_{2n})=3^n+1\]
Thus $s=\frac{3^n+1}{2}$, or $\boxed{E}$.
~ Nafer
~ Minor edits by Qinglang, Nafer wrote a great solution, but put D instead of E