binomial theorem!

In theexpansion of [ax+(2/x^2)]^n，the 3rd term in descending powersof x is (15/4)，
where n is a positive integer and a<0，Find the values of n and a.
Sol
[ax+(2/x^2)]^n
=(ax)^n+c(n，1)*(ax)^(n－1)*(2/x^2)^1+c(n，2)*(ax)^(n－2)*(2/x^2)^2+….
So
c(n，2)*(ax)^(n－2)*(2/x^2)^2=15/4
x^(n－2)*x^(－4)=x^0
n－6=0
n=6
c(6，2)*a^(6－2)*2^2=15/4
15*a^4*4=15/4
a^4=1/16
a<0
a=－1/2