Answer
$${\text{twelfth term}} = 139776{x^5}{y^{22}}$$
Work Step by Step
$$\eqalign{
& {\text{twelfth term of }}{\left( {2x + {y^2}} \right)^{16}} \cr
& {\text{In the twelfth term, }}{y^2}{\text{ has an exponent of 12}} - 1,{\text{ or 11, }} \cr
& {\text{while }}2x{\text{ has an exponent of 16}} - 11,{\text{ or 5}}{\text{.}} \cr
& \cr
& {\text{Using the }}k{\text{th Term of The Binomial Expansion}} \cr
& {\text{twelfth term}} = \frac{{16!}}{{5!11!}}\left( {32{x^5}} \right)\left( {{y^{22}}} \right) \cr
& {\text{twelfth term}} = 4368\left( {32{x^5}} \right)\left( {{y^{22}}} \right) \cr
& {\text{twelfth term}} = 139776{x^5}{y^{22}} \cr} $$
