Given $$P(x)=(1+x+x^2)^{100}=a_0+a_1x+\cdots+a_{200}x^{200}$$
Compute the following sums:
Setting $x=1$ gives the result for $S_1$: $$S_1=a_0+a_1+a_2+\cdots + a_{200} = \boxed{3^{100}}$$
Next, setting $x=-1$ yields $$S_{-1} = a_0 - a_1 + a_2 +\cdots + a_{200} = 1^{100} = 1$$
Now, we have $$S_2 = \frac{1}{2}\cdot\left(S_1 + S_{-1}\right)=\boxed{\frac{3^{100} + 1}{2}}$$