Find the value of three expressions

Question

Consider $$(1+x+x^2)^n=\sum_{r=0}^{2n}a_rx^r$$ where $a_0,a_1,a_2,\cdots,a_{2n}$ are real numbers and $n$ is a positive integer.

Then find the value of

$$\sum_{r=0}^{n-1}a_{2r}$$

And

$$\sum_{r=1}^na_{2r-1}$$

And

$$a_2$$

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My work:

I expanded $(1+x+x^2)^n$ as $$\sum_{i+j+k=n}\binom{n}{i,j,k}1^ix^jx^{2k}$$
But i don't think that it's any useful to us. Then I tried putting $x=1,\omega,\omega^2$ but then I got,
$$3^{n-1}=a_0+a_3+\cdots$$

Any help is greatly appreciated.

Answer 1

If $$ (1+x+x^2)^n=\sum_{k=0}^{2n}a_kx^k, $$ then $$ 3^n=(1+1+1^2)^n=\sum_{k=0}^{2n}a_k\quad\text{and}\quad 1=(1-1+1^2)^n=\sum_{k=0}^{2n}(-1)^ka_k. $$ Hence $$ \sum_{k=0}^{n}a_{2k}=\frac{1+3^n}{2}\quad\text{and}\quad \sum_{k=1}^{n}a_{2k-1}=\frac{3^n-1}{2} $$ and if $f(x)=(1+x+x^2)^n=\sum_{k=0}^{2n}a_kx^k$, then $f''(0)=2a_2$. But $$ f''(x)=2n(1+x+x^2)^{n-1}+n(n-1)(1+2x)^2(1+x+x^2)^{n-2} $$ and finally $$ 2a_2=2n+n(n-1)=n(n+1). $$