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.
On
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.
On
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.
$$\left(x+\frac{1}{x}+x^2 + \frac{1}{x^2}\right)^{15}$$ $$= \left(\frac{x^{3}+ x + x^4 + 1}{x^{2}}\right)^{15}$$ $$= \left(\frac{1 + x+ x^3 + x^4}{x^{2}}\right)^{15}$$ $$= x^{-30} \cdot\left(1 + x+ x^3 + x^4\right)^{15}$$
Consider the expansion of $\left(1 + x+ x^3 + x^4\right) ^ {15}$.
We have the following characteristics:
The 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.