If $$(1 + x + x^2)^n = \sum^{2n}_{k=0} b_k x^{2k}$$
Then find $\sum^{2n}_{k= 0} b_k$
$$((1+ x) + x^2)^n = \sum^n_{k=0}{n\choose k} x^{2k}(1+x)^{n-k} = \sum^n_{k=0}{n\choose k} x^{2k}\sum^{n-k}_{r = 0} {n-k \choose r} x^r = \sum^n_{k=0}\sum^{n-k}_{r = 0}{n\choose k} x^{2k} {n-k \choose r} x^r $$
$${n\choose k}{n-k \choose r} = \dfrac{n!}{k! (n-k)!}\dfrac{(n-k)!}{k! (n-k - r)!} = \dfrac{n!}{(k!)^2 (n-k-r)!}$$
So,
$$\sum^n_{k=0}\sum^{n-k}_{r = 0}{n\choose k} x^{2k} {n-k \choose r} x^r = \sum^n_{k=0}\sum^{n-k}_{r = 0} x^{2k+r}\dfrac{n!}{(k!)^2 (n-k-r)!}$$
I don't have much idea what to do now to do now. Is the question even correct ?