Extract sum of coefficients in a binomial expression

Question

I have two questions:

(1) Given $(1-x+x^2)^{3n}=c_0 + c_1 x + \dots +c_{6n} x^{6n}$, find $c_0+c_1+ \dots +c_n$.

I manage to find $c_0+c_1+ \dots +c_{6n}$ by putting $x=1$ but I do not know how to find the required sum up to $n$. To me, it seems that it omits terms from $n+1^{th}$ to $6n^{th}$ and thus I do not know how to find it properly.

(2) Futher given $(1+x)^{3n}=k_1 +k_1 x + \dots k_{3n} x^{3n}$, find $c_0 b_0 + c_1 b_1 +\dots c_{3n}b_{3n}$.

I have no idea on how this is related to question (1).

Thanks in advance.

Answer 1

Set $x=1/y$ in the first relation and multiply by $y^{6n}$ to get

$$(1-y+y^2)^{3n}=c_0y^{6n}+c_1y^{6n-1}+\cdots+c_{6n}$$

Set $x=y$ in the second to get $(1+y)^{3n}=k_0+k_1y+k_2y^2+\cdots+k_{3n}y^{3n}$

So, $\sum_{r=0}c_rk_r$ is the coefficient of $y^{6n}$ in

$$(1-y+y^2)^{3n}\cdot(1+y)^{3n}=\left[(1-y+y^2)(1+y)\right]^{3n}=(1+y^3)^{3n}$$

Now the $r$th term of $(1+y^3)^{3n}$ is $$\binom{3n}r(y^3)^r=\binom{3n}ry^{3r}$$

So, we need $3r=6n\iff r=?$