Question

Find the third term in the expansion of ${(2x^2+\frac{3}{2x})}^8$

Answer 1

${(2x^2+\frac{3}{2x})}^8$

${=}^8{c}_2(2x^2{)}^6{\left(\frac{3}{2x}\right)}^2$

$=\frac{8!}{6!\times 2!}\times 2^6\times x^{12}\times \frac{9}{4\times x^2}$

$=4032x^{10}$