# binomial theorem: find coef. xy

Given:

$$\left(x-\dfrac{1}{2y}\right)^8\left(x+\dfrac{1}{2y}\right)^4$$

Using binomial theorem, what is the coefficient of xy in the expansion?

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 I converted the expression to $\LaTeX$. Is this what you meant? – Ayman Hourieh Sep 17 '12 at 8:40 yes is it, thank you – David Hoffman Sep 17 '12 at 20:47

If you multiply $$\left( \sum_{k=0}^8 \binom{8}{k}{x^k {\left(-1 \over 2y\right)}^{8-k} }\right) \left( \sum_{l=0}^4 \binom{4}{k}{x^l {\left(1 \over 2y\right)}^{4-l} }\right)$$ you will never get $xy$
$y$ is in denominator in both the terms $\left(x-\dfrac{1}{2y}\right)^8$ and $\left(x+\dfrac{1}{2y}\right)^4$ and powers are $4$ and $8$(both positive). So, how do you expect to get $xy$.
Simply, the coefficient of $xy$ is $0$