A simple yet tricky binomial question

Question

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

I know how to solve this kind of problem.

First I would arrange the term into a binomial expression. The expansion will have $(n+1)$ dissimilar terms.

But how can I arrange it into a binomial expression?

Answer 1

Hint :

$$x+\dfrac{1}{x} = t \Rightarrow x^2+\dfrac{1}{x^2} = t^2 - 2$$

$$\therefore \Big(x+\dfrac{1}{x}+x^2+\dfrac{1}{x^2}\Big)^{15} = (t^2+t-2)^{15} $$

Now consider @lulu's precious advice : "What's the highest degree term? What's the lowest? Do all the intermediate terms have non-zero coefficients?"

Answer 2

This is how I would proceed with the question:

$(x+\frac{1}{x}+x^2+\frac{1}{x^2})^{15}$ = $(\frac{1+x+x^3+x^4}{x^2})^{15}$ = $(\frac{1}{x^30})(1+x+x^3+x^4)^{15}$ = $(\frac{1}{x^30})$(1+.......+x^60)

That expression has 61 different powers. So the answer should be 61. Hope it helps!