<s>[INST] Expression of $x^n+\frac1{x^n}$ by $x+\frac1{x}$ where $n$ is a positive odd number. There was a problem in a book:

Denote that $y=x+\dfrac{1}{x}$, express $x^7+\dfrac{1}{x^7}$ using $y$.

It's not a hard question, but I find a special sequence:
$x+\dfrac{1}{x}=y\\x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)=y^3-3y \\ x^5+\dfrac{1}{x^5}=\left(x+\dfrac{1}{x}\right)^5-5\left(x^3+\dfrac{1}{x^3}\right)-10\left(x+\dfrac{1}{x}\right)=y^5-5\left(y^3-3y\right)-10y=y^5-5y^3+5y\\x^7+\dfrac{1}{x^7}=\left(x+\dfrac{1}{x}\right)^7-7\left(x^5+\dfrac{1}{x^5}\right)-21\left(x^3+\dfrac{1}{x^3}\right)-35\left(x+\dfrac{1}{x}\right)=y^7-7\left(y^5-5y^3+5y\right)-21\left(y^3-3y\right)-35y=y^7-7y^5+14y^3-7y$
I find that the coefficient has some relationship between the Pascal Triangle, such as $y^7-7y^5+14y-7y=y^7-7 \binom{2}{0}y^5+7\binom{2}{1}y^3-7\binom{2}{2}y$. That's strange but funny! However, I can't really prove this, or show that it is false. Hope there is someone who can answer me. Thank you!
 [/INST] Krechmar's `A problem book in Algebra' gives an identity if $x+z=p$ and $xz=q$, then
$$x^n+z^n=p^n-\frac{n}{1}p^{n-2} q+\frac{n(n-3)}{1.2}p^{n-4} q^2+...+(-1)^k \frac{n(n-k-1)(n-k-2)....(n-2k+1)}{k!} p^{n-2k} q^k+...$$
Take $z=1/x$ then $q=1$ and letting $x+\frac{1}{x}=y$ we can write 
$$f_n(x)=x^n+\frac{1}{x^n}=\sum_{k=0}^{n/2} A_{n,k} ~y^{n-2k},...(1)$$
where $$A_{n,k}=(-1)^k \frac{n(n-k-1)(n-k-2)....(n-2k+1)}{k!}....(2)$$
We can re-write 
$A_{n,0}=1$ and $$A_{n,k}=(-1)^k\frac{n}{k}{n-k-1 \choose k-1}, ~ 0<k <n/2 ......(3)$$
Fpr $n=5$ we get $A_{5,0}=1, A_{5,1}= -5, A_{5,2}=5.$
For $n=7$ we get $A_{7,0}=1, A_{7,1}=-7, A_{7,2}= 14, A_{7,3}=-7$
For $n=8$, we get $A_{8,0}=1, A_{8,1}=-8, A_{8,2}=  20, A_{8,3}=-16, A_{8,4}=2.$
So $$x^8+1/x^8=y^8-8y^6+20y^4-16y^2+2,~ y=x+1/x.$$
Similarly $$x^9+1/x^9=y^9-9y^7+27y^5-30y^3+9y.$$
Finally, we would like to assert that the expansion (1) along with (2) or (3) is the required generalization which works for both even and odd $n$.
</s>