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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. |
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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. |
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Perhaps note that $$((x^2-1)/x)^{100} = x^{-100} \sum_{j=0}^{100} \binom{100}{j} x^{2j} (-1)^j = \sum_{j=0}^{100} \binom{100}{j} (-1)^j x^{2j-100}.$$ 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 $0$. |
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