Answer
$ T_1=x^8$
$ T_2=8x^7 y^2$
$ T_3=28x^6 y^4$
Work Step by Step
The formula to compute $ T_{k+1}$ of the expansion is: $ T_{k+1} =\dbinom{n}{k}p^{n-k}q^k $
Given: $(x+y^2)^8$
$ n=8; k=0$
$ T_1=\dbinom{8}{0}x^{8}(y^2)^0=x^8$
$ n=8; k=1$
$ T_2=\dbinom{8}{1}x^{7}(y^2)^1=8x^7 y^2$
$ n=8; k=2$
$ T_3=\dbinom{8}{2}x^{6}(y^2)^2=28x^6 y^4$
