Answer
$16A^{4}+32A^{3}B^{2}+24A^{2}B^{4}+8AB^{6}+B^{8}$
Work Step by Step
$(2A+B^{2})^{4}= \,^4C_0\times(2A)^{4}\times(B^{2})^{0}+\,^4C_1\times(2A)^{3}\times(B^{2})^{1}+\,^4C_2\times(2A)^{2}\times(B^{2})^{2}+\,^4C_3\times(2A)^{1}\times(B^{2})^{3}+\,^4C_4\times(2A)^{0}\times(B^{2})^{4}$
$=(1\times16\times A^{4}\times1)+(4\times8\times A^{3}\times B^{2})+(6\times4A^{2}\times B^{4})+(4\times2A\times B^{6})+(1\times1\times B^{8})$
$=16A^{4}+32A^{3}B^{2}+24A^{2}B^{4}+8AB^{6}+B^{8}$
