I am trying to find the constant term in this expansion:
$$\bigg(1 + x + 2y^{2} - \frac{1}{x^{2}y}\bigg)^{15}$$
I have been trying for hours but I hit a wall when working out the values $(r_1, r_2, r_3, r_4)$. I get $2$ equations $r_3 = r_2/4$ and $r_4 = r_2/2$ and can't get any further. maybe I'm just being stupid but any help is greatly appreciated
$$\bigg(1 + x + 2y^{2} - \frac{1}{x^{2}y}\bigg)^{15}=\sum\frac{15!}{(n_1)!(n_2)!(n_3)!(n_4)!}(1)^{n_1}(x)^{n_2}(2y^2)^{n_3}\Bigl(\frac{-1}{x^2y}\Bigr)^{n_4}$$ $$=\sum\frac{15!}{(n_1)!(n_2)!(n_3)!(n_4)!}(2)^{n_3}(-1)^{n_4}(x)^{n_2-2n_4}(y)^{2n_3-n_4}$$ $$2n_3-n_4=0$$ $$n_2-2n_4=0$$ $$n_2=4n_3=2n_4$$ $$n_1+n_2+n_3+n_4=15$$ $$n_1+4n_3+n_3+2n_3=15$$ $$n_1+7n_3=15$$ $$(n_1,n_3)=(15,0),(8,1),(1,2)$$ Thus, $$(n_1,n_2,n_3,n_4)=(15,0,0,0) , (8,4,1,2), (1,8,2,4)$$
$$C=\sum\frac{15!}{(15)!(0)!(0)!(0)!}(2)^{0}(-1)^{0}(x)^{0-0}(y)^{0-0}+\sum\frac{15!}{(8)!(4)!(1)!(2)!}(2)^{1}(-1)^{2}(x)^{4-2*2}(y)^{2*1-2}+\sum\frac{15!}{(1)!(8)!(2)!(4)!}(2)^{2}(-1)^{4}(x)^{8-2*4}(y)^{2*2-4}$$ $$C=1+\frac{2\times 15! }{(8!)(2!)(4!)}+\frac{4\times 15! }{(8!)(2!)(4!)}=1+\frac{6\times 15! }{(8!)(2!)(4!)}=4054051$$