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The answer in the textbook in the solutions manual is the following:

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I didn't undestand the last step. He said he did a binomial expansion, I just can't figure out! Appreciate any help.

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The binomial expansion says that $(1+x)^n=1+{n \choose 1}x^1+{n \choose 2}x^2 + ...$. This should be familiar to you for positive, integer n just by expanding out the parenthesis. For NEGATIVE n, it still holds, provided you interpret ${n \choose k}$ correctly for negative numbers; for our purposes, we just need to know ${n\choose 1}=n$ always.

For very small $x$, the higher-order terms are negligable, so we can just say $$ (1+x)^n = 1+{n \choose 1}x^1+{n \choose 2}x^2 + ... \approx 1+nx $$

If you rewrite $(z-d/2)^{-3}$ as $z^{-3}(1-d/2z)^{-3}$ and then follow your nose, you should get the answer quoted.

Note: You can DERIVE this formula for negative (or even fractional) $n$ just by considering the Taylor expansion of $(1+x)^n$.

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