Coefficient of $x^{12}$ in $(1+x^2+x^4+x^6)^n$ I need to find the coefficient of $x^{12}$ in the polynomial $(1+x^2+x^4+x^6)^n$.
I have reduced the polynomial to $\left(\frac{1-x^8}{1-x^2}\right)^ n$ and tried binomial expansion and Taylor series, yet it seems too complicated to be worked out by hand.
What should I do?
 A: $(1+x^2+x^4+x^6)^n = (1+x^4)^n(1+x^2)^n$
the $r^{th}$ term in first series: ${n \choose r} x^{4r}$
the $l^{th}$ term in second series: ${n \choose l} x^{2l}$
so we have $4r+2l = 12 \implies r=0, l=6; r=1, l=4; r=2, l=2; r=3, l=0$
So we have coefficient:  ${n \choose 0}{n \choose 6} +{n \choose 1}{n \choose 4} + {n \choose 2}{n \choose 2} + {n \choose 3}{n \choose 0}$
A: Your expression equals $$(1+x^2)^n (1+x^4)^n.$$ You should be able to finish using the binomial theorem.
A: You can try using the multinomial theorem
$$(1+x^2+x^4+x^6)^n = \sum_{0 \le a,b,c,d \le n} \dfrac{n!}{a!~b!~c!~d!}~1^a~x^{2b}~x^{4c}~x^{6d}  $$
$$ a+b+c+d=n ~~~~;~~~ a,b,c,d \in \mathbb{W}$$
For the coefficient of $x^{12}$,
$$ 2b+4c+6d = 12$$
$$  b +2c+3d=6$$
Also, $$0 \le a,b,c,d \le n $$
$$0 \le b \le6 $$
$$0 \le c \le3 $$
$$0 \le d \le2 $$
Now you just have to find all solutions that satisfy the above criteria.
A: Here is just a standard trick from generating functions using
$$\frac 1{(1-y)^n} = \sum_{k=0}^{\infty}\binom{k+n-1}{n-1}y^k$$.
To simplify the expressions set
$$y=x^2\Rightarrow \text{ we look for }[y^6]\frac{(1-y^4)^n}{(1-y)^n}$$
Hence, for $n\geq 1$ you get
\begin{eqnarray*}[y^6]\frac{(1-y^4)^n}{(1-y)^n}
& = & [y^6]\left((1-y^4)^n\sum_{k=0}^{\infty}\binom{k+n-1}{n-1}y^k\right) \\
& = & [y^6]\left((1-ny^4)\sum_{k=0}^{\infty}\binom{k+n-1}{n-1}y^k\right) \\
& = & \boxed{\binom{6+n-1}{n-1} - n\binom{2+n-1}{n-1}}
\end{eqnarray*}
A: Your approach is also fine. In the following it is convenient to denote with $[x^k]$ the coefficient of $x^k$ in a series.

We obtain
\begin{align*}
\color{blue}{[x^{12}]}&\color{blue}{\left(\frac{1-x^8}{1-x^2}\right)^n}\\
&=[x^{12}](1-x^8)^n\sum_{j=0}^{\infty}\binom{-n}{j}\left(-x^2\right)^j\tag{1}\\
&=[x^{12}]\left(1-\binom{n}{1}x^8\right)\sum_{j=0}^{\infty}\binom{n+j-1}{j}x^{2j}\tag{2}\\
&=\left([x^{12}]-n[x^4]\right)\sum_{j=0}^{\infty}\binom{n+j-1}{j}x^{2j}\tag{3}\\
&\,\,\color{blue}{=\binom{n+5}{6}-n\binom{n+1}{2}}\tag{4}
\end{align*}

Comment:

*

*In (1) we use the binomial series expansion.


*In (2) we expand $(1-x^8)^n$ up to terms of $x^8$, since other terms do not contribute. We also use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.


*In (3) we apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$.


*In (4) we select the coefficients of $x^k$ accordingly.
