Coefficient of $x^{12}$ in $(1+x^2+x^4+x^6)^n$

Question

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?

Answer 1

$(1+x^2+x^4+x^6)^n = (1+x^4)^n(1+x^2)^n$

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: ${n \choose 0}{n \choose 6} +{n \choose 1}{n \choose 4} + {n \choose 2}{n \choose 2} + {n \choose 3}{n \choose 0}$

Answer 2

Your expression equals $$(1+x^2)^n (1+x^4)^n.$$ You should be able to finish using the binomial theorem.

Answer 3

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.

Answer 4

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*}

Answer 5

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.