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?
I've tried to do it but I couldn't. Could you please help me with it?
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?
I've tried to do it but I couldn't. Could you please help me with it?
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 $0$