To find number of distinct terms of binomial expansion I have binomial expansion as 
$(x+\frac{1}{x} + x^2 + \frac{1}{x^2})^{15}$. 
How do i find number of distinct terms in it. Distinct in sense means terms having different powers of $x$?
I have simplified this as $\frac{1}{x^{30}}(x^3+x+x^4+1)^{15}$. How do i proceed
Thanks

We have that
$$(x^4+x^3+x+1)^n=(x+1)^n\cdot(x^3+1)^n.$$
Now the powers of $x$ in $(1+x)^n$ are the integers in the interval $[0,n]$.
Moreover for $n\geq 2$, the multiplication of $(x^3+1)(x+1)^n$ gives a sum of powers of $x$ which cover the range $[0,n]\cup [3,n+3]=[0,n+3]$. 
Hence, for $(x+1)^n\cdot(x^3+1)^n$ the powers of $x$ cover the range $[0,n+3n]=[0,4n]$, which contains $4n+1$ integers. 
Therefore
$$\left(x+\frac{1}{x} + x^2 + \frac{1}{x^2}\right)^{15}=\frac{1}{x^{30}}(x^3+x+x^4+1)^{15}$$
contains all the powers of $x$ in the range $[-30,4\cdot 15-30]=[-30,30]$. Their number is $61$.
