# 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 $x^k$

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No, $(\frac {x^2-1}x)^{100} \ne \frac {x+1}x$. You lost the power $100$ and a factor $x-1$ from the numerator. Also please see meta.math.stackexchange.com/questions/1773/… on how to format equations in $\LaTeX$. They are much easier to read that way. –  Ross Millikan Feb 21 '13 at 15:24
whoops, made the clarification –  user60862 Feb 21 '13 at 15:26
It still doesn't simplify that way. $x^2-1 \ne x+1$ –  Ross Millikan Feb 21 '13 at 15:27
Then the sum is wrong. You have $(\frac {x+1}x)$, but this is $1+\frac 1x$, not $x+\frac 1x$ which is what you would need for the sum. –  Ross Millikan Feb 21 '13 at 15:31

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.
$$((x^2-1)/x)^{100} = x^{-100} \sum_{j=0}^{100} \binom{100}{j} x^{2j} (-1)^j = \sum_{j=0}^{100} \binom{100}{j} (-1)^j x^{2j-100}.$$
Therefore, if you would like the coefficient of $x^k$, set $k = 2j-100$ and note that $j = (k+100)/2 = 50 + k/2$ and so for even $k$ between $-100$ and $100$, the coefficient is $\binom{100}{50+k/2} (-1)^{k/2}$ and for odd $k$ it is $0$.