Number of dissimilar terms in the expansion of given multinomial

Question

The number of dissimilar terms in the expansion of $$\bigg( x + \frac{1}{x}+x^2+\frac{1}{x^2}\bigg)^{15} $$ are:

Using binomial theorem, $$\bigg( x + \frac{1}{x}+x^2+\frac{1}{x^2}\bigg)^{15} $$ $$=\sum_{i=0}^{15} \binom{15}{i}\bigg(x+\frac{1}{x}\bigg)^{i}\bigg(x^2+\frac{1}{x^2}\bigg)^{15-i}$$ But what to do next! Please solve this question.

Answer 1

Hint:  $\dfrac{1}{x^{30}}(x^4+x^3+x^2+x+1)^{15}\,$ has all the terms in $\,x^k\,$ non-zero for $\,-30 \le k \le 30\,$.

Answer 2

Note that if we square the given expression, namely: $\left(x+\frac1{x} + x^2 + \frac1{x^2} \right)$, five new terms: $\frac1{x^4}, \frac1{x^3},1,x^3$ and $x^4$ are introduced. Thereafter, successive multiplication by yields only four new terms which can be checked by computation.

Starting from $n=2$, the total number of distinct terms forms an arithmetic sequence whose $n^{\text{th}} $ term is given by: $$ a_n=4n+1$$

Hence, with $n=15$, we have $61$ distinct terms.