Formula for $x^k$ in $((x^2 - 1)/x)^{100}$ So $((x^2 - 1)/x)^{100}$  simplifies to $((x+1)/x)^{100}$
Through the binomial formula $((x+1)/x)^{100}$ is:
$$\sum_{j = 0}^{100} C(100,j) x^{100-j} (1/x)^j$$
by simplification this is:
$$\sum_{j = 0}^{100} C(100,j) x^{2j - 100}$$
I'm not sure how to proceed and find the formula for $x^k$
Help please.

For every $0\leqslant k\leqslant100$, the coefficient of $x^{100-2k}$ in $((x^2 - 1)/x)^{100}$ is $(-1)^k{100\choose k}$. These are the even powers between $-100$ and $+100$. The coefficient of every other power is zero.
