Question

Find the $7^{th}$ term in ${(\frac{4}{x^3}+\frac{x^2}{2})}^{14}$

Answer 1

${(\frac{4}{x^3}+\frac{x^2}{2})}^{14}$

$T_7={}^{14}{C}_6{\left(\frac{4}{x^3}\right)}^8{\left(\frac{x^2}{2}\right)}^6$

$=\frac{14!}{6!8!}\times \frac{4^8}{x^{24}}\times \frac{x^{12}}{2^6}$

$=\frac{14\times 13\times 12\times 10\times 9}{6\times 5\times 3\times 2}\times \frac{2^{16}}{2^6}\times \frac{1}{x^{12}}$

$=\frac{14\times 13\times 6\times 2^{10}}{x^{12}}$

$=\frac{1118208}{x^{12}}$.