If $(1+x)^{4n} +(1+x+x^2)^{2n} +(1+x+x^2+x^3+x^4)^n = a_0 + {a_1}x + {a_2} x^2 + .... +{a_{4n}}x^{4n}$ , then prove that $a_r=a_{4n-r}$ I tried solving this question by attempting to prove $(a_r)=(a_{4n-r})$
Now, $a_r$ is the coefficient of $x^r$ which we can obtain by adding up the coefficients of $x^r$ from the 3 separate expressions $(1+x)^{4n}$ , $(1+x+x^2)^{2n}$ and $(1+x+x^2+x^3+x^4)^n$
In the first expression $(1+x)^{4n}$, it is fairly easy to see that the coefficient of $x^r$ can be obtained from binomial expansion and the coefficient $a_r=a_{4n-r}$ by some basic properties.
I tried finding the coefficient of $x^r$ in $(1+x+x^2)^{2n}$
I first tried doing so by simplifying the expression as
$$(1+x+x^2)^{2n} = [1+(x(1+x)]^{2n}$$
Then taking $x(1+x)$ as some $y$, I applied binomial expansion and obtained the following expansion,
$$[1+(x(1+x)]^{2n} = \binom {2n}{r}\binom{r}{0}+\binom{2n}{r-1}\binom{r-1}{1}+\binom{2n}{r-2}\binom{r-2}{2}+...+\binom{2n}{r/2}\binom{r/2}{r/2}$$
(when $r$ is even, otherwise instead of going upto $r/2$ we will go upto $(r-1)/2$)
However, after calculating this, I was unable to find any relation between $a_r$ and $a_{4n-r}$ for this expression.
I was also not able to find any such simplification for the third expression $(1+x+x^2+x^3+x^4)^n$
I tried using the identity
$$1+x+x^2+...+x^n = {(1-x^{n+1})}/{(1-x)}$$
But was unable to obtain any further simplification even on using negative binomial expansions.
Any help on how to approach this question is appreciated
Thanks in advance!
Regards

Let $$P(x)=(1+x)^{4n} +(1+x+x^2)^{2n} +(1+x+x^2+x^3+x^4)^n$$ Then it's reciprocal polynomial is $$\hat{P}(x)=x^{4n}P(1/x)=x^{4n}\left(\left(1+\frac{1}{x}\right)^{4n} +\left(1+\frac{1}{x}+\frac{1}{x^2}\right)^{2n} +\left(1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^4}\right)^n\right)=P(x)$$ Hence $P(x)$ is a palindromic polynomial and hence $a_r=a_{4n-r}$.
