## Precalculus (6th Edition) Blitzer

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