I am trying to find the coefficients of the following two terms in expansion of $(3x-y^2)^{44}$.
For the first one we need to find the coefficients of $x^{20}y^{48}.$
I found this one by following formula $(x+y)^n = \sum_{k=0}^n \binom{n}{k}x^{n-k}y^{k}$ $$(3x-y^2)^{44} = \sum_{k=0}^{44} \binom{44}{k}(3x)^{44-k}(-y)^{2k} \\ \text{Let $k=24$} \\ = \sum_{k=0}^{44} \binom{44}{24}(3x)^{44-24}(-y)^{2(24)} \\ (3x-y^2)^{44}= \binom{44}{24}(3x)^{20}(-y)^{48}$$ Now I am trying to find the coefficients for $x^{28}y^{30}$.
$$(3x-y^2)^{44} = \sum_{k=0}^{44} \binom{44}{k}(3x)^{44-k}(-y)^{2k}$$ But I am stuck here becauseI don't know what $k$ should equal to have the coefficients to be $x^{28}y^{30}$. Is there a better way to find this $k$?
The coefficient of $x^{28}y^{30}$ must be zero, we get two contradictory values of $k=16$ and $k=15$.