Number of terms in $\left(x+\dfrac{1}{x}+x^2 + \dfrac{1}{x^2}\right)^{15}$ Find the number of terms in $$\left(x+\frac{1}{x}+x^2 + \frac{1}{x^2}\right)^{15}$$
I dont have any clue on how to start this problem. Terms can be formed from various combinations! Thank you!
Edit
Actually answer provided is $61$, so it seems all powers are present.
 A: $$\left(x+\frac{1}{x}+x^2 + \frac{1}{x^2}\right)^{15}$$
$$= \left(\frac{x^{3}+ x + x^4 + 1}{x^{2}}\right)^{15}$$
$$= \left(\frac{1 + x+ x^3 + x^4}{x^{2}}\right)^{15}$$
$$= x^{-30} \cdot\left(1 + x+ x^3 + x^4\right)^{15}$$
Consider the expansion of $\left(1 + x+ x^3 + x^4\right) ^ {15}$.
We have the following characteristics:


*

*The maximum power of $x$ in the expansion is $60$

*Each term will have a power which is formed as follows: let there be 15 slots and we fill them with numbers from the set $\{0, 1, 3, 4\}$ which are the powers of $x$ in the expression. Can we construct sets such the sum of numbers in each set will be able to represent all numbers from 0..60?


The answer to (2) seems to be in the affirmative. The proof is a simple exercise. Consider all of the $15$ slots to be filled with the value $4$. That will give rise to $60$. If you want to have construct numbers from $56 - 59$, it can be done as follows:
$$60: 15 \cdot 4 + 1         : 15\, slots$$
$$59: 14 \cdot 4 + 1 \cdot 3 : 15\, slots$$
$$58: 13 \cdot 4 + 2 \cdot 3 : 15\, slots$$ 
$$57: 14 \cdot 4 + 1 \cdot 1 : 15\, slots$$ 
$$56: 14 \cdot 4 + 1 \cdot 0 : 15\, slots$$
We can provide a simple inductive proof based on the above reasoning. Hence we will have all powers in the expansion from $\{0..60\}$ and hence the answer is $\boxed{61}$.
Side Note: This problem is quite similar to a problem on weights, and we could be sure that in such cases where we have a good set of weights and a large enough number of slots, we could represent a large number of values. However, if you consider weights of the form $\{1, 3, 5\}$ and you want to represent values from $\{0..15\}$, i.e., the numbers in the expansion of $\left(x + x^3 + x^9\right)^3$ will not have all values (since $4$ for example is not representable as such a sum.
A: The "top term" will be $x^{30}$ and the "bottom term" will be $x^{-30}$.
In between there will be terms involving $x^{29}$, $x^{28}$ etc., with
various coefficients.
There seems no reason to think any coefficients will vanish.
A: We have
$$ f(x)=(x^{-2}+x^{-1}+x^{1}+x^{2})^{15} = \sum_{k=-30}^{30}c_k x^k \tag{1}$$
where $c_k$ is given by the number of ways for representing $k$ as $-2a-b+c+2d$ with $a,b,c,d\in\mathbb{N}$ an $a+b+c+d=15$. We clearly have $c_k=c_{-k}$ by swapping $a$ and $d$, $b$ and $c$. By direct inspection every coefficient of $g(x)=(x^{-2}+x^{-1}+x^{1}+x^{2})^{3}$ is positive, hence the same holds for the coefficients of $f(x)=g(x)^5$. This gives
$$ \forall k\in[-30,30],\qquad c_k > 0 \tag{2}$$
hence there are $61$ terms.

In general, if $a_1,a_2,\ldots,a_{n}$ with $n\geq 2$ is a sequence of distinct integers, the Laurent series of
$(x^{a_1}+x^{a_2}+\ldots+x^{a_n})^m $ has positive coefficients for any $m\in\mathbb{N}$ large enough.
