Find the number of dissimilar terms in the expansion of $$\bigg(x+\frac{1}{x}+x^2+\frac{1}{x^2} \bigg)^{20}$$
Could someone give me some hint to proceed in this question. I used $x+1/x=t$ and wrote expansion as $(t^2+t-2)^{20}$ but even this does not convey much information. Please help.
On
Here are some additional hints.
Applying the binomial theorem we obtain \begin{align*} \left(x+\frac{1}{x}+x^2+\frac{1}{x^2} \right)^{20}=\sum_{j=0}^{20}\binom{20}{j}\left(x+\frac{1}{x}\right)^j\left(x^2+\frac{1}{x^2}\right)^{20-j}\tag{1} \end{align*}
Since \begin{align*} \left(x+\frac{1}{x}\right)\left(x^a+\frac{1}{x^a}\right)=\left(x^{a+1}+\frac{1}{x^{a+1}}\right)+\left(x^{a-1}+\frac{1}{x^{a-1}}\right) \end{align*} and \begin{align*} \left(x^2+\frac{1}{x^2}\right)\left(x^a+\frac{1}{x^a}\right)=\left(x^{a+2}+\frac{1}{x^{a+2}}\right)+\left(x^{a-2}+\frac{1}{x^{a-2}}\right) \end{align*}
we conclude from (1) \begin{align*} \left(x+\frac{1}{x}+x^2+\frac{1}{x^2} \right)^{20}=\sum_{j=0}^{40} a_j\left(x^{j}+\frac{1}{x^j}\right) \end{align*}
Argue that $a_j\ne 0$ where $0\leq j\leq 40$.
It follows the expression has $81$ dissimilar terms.
When you square $\left(x^{-2}+x^{-1}+x+x^2\right)$ you get five new terms, namely $x^{-4},x^{-3},1,x^3,x^4$. Thereafter, multiplication by $\left(x^{-2}+x^{-1}+x+x^2\right)$ yields only four new terms.
The cube introduces the four new terms $\left(x^{-4}+x^{-3}+x^3+x^4\right)$, the fourth power introduces the four new terms $\left(x^{-6}+x^{-5}+x^5+x^6\right)$, etc.
So starting with $n=2$ the total number of distinct terms forms an arithmetic sequence
$$ a_n=4n+1 \text{ for } n>1$$
Thus $\left(x^{-2}+x^{-1}+x+x^2\right)^{20}$ will have $4(20)+1=81$ distinct terms.
Note: One could also rewrite it as $\left(\dfrac{1+x+x^3+x^4}{x^2}\right)^{20}$ which would be a degree $80$ polynomial divided by $x^{40}$ which gives a maximum of $81$ terms, but one would have to demonstrate that the missing $x^2$ term does not reduce that number.