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

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

I've tried to do it but I couldn't. Could you please help me with it?

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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
up vote 2 down vote accepted

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 $xy$

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$y$ 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 $0$

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