<s>[INST] Number of dissimilar terms in given expansion 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.
 [/INST] 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.
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