Answer
$${\text{seventeenth term}} = 74613{a^{12}}{b^{16}}$$
Work Step by Step
$$\eqalign{
& {\text{seventeenth term of }}{\left( {{a^2} + b} \right)^{22}} \cr
& {\text{In the seventeenth term, }}b{\text{ has an exponent of 17}} - 1,{\text{ or 16, }} \cr
& {\text{while }}{a^2}{\text{ has an exponent of 22}} - 16,{\text{ or 6}}{\text{.}} \cr
& \cr
& {\text{Using the }}k{\text{th Term of The Binomial Expansion}} \cr
& {\text{seventeenth term}} = \left( {22{\bf{C}}16} \right){\left( {{a^2}} \right)^6}{\left( b \right)^{16}} \cr
& {\text{seventeenth term}} = \frac{{22!}}{{6!16!}}{\left( {{a^2}} \right)^6}{\left( b \right)^{16}} \cr
& {\text{seventeenth term}} = 74613{a^{12}}{b^{16}} \cr} $$
