Question

Find the middle term (s) in the expansion of the following
${(x^{\frac{3}{2}}-y^{\frac{3}{2}})}^8$

Answer 1

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

$\because n=8$ is even so

$\therefore$ No. of term $=n+1=9$

So middle term $=5th$ term

$T_5=T_{4+1}{=}^9{C}_4{\left(x^{\frac{3}{2}}\right)}^{9-4}{(-y^{\frac{3}{2}})}^4$

$=\frac{9!}{5!4!}{\left(x^{\frac{3}{2}}\right)}^5(-y^{\frac{3}{2}\times 4})$

$=\frac{9\times 8\times 7\times 6\times 5!}{5!4!\times 3\times 2}{x}^{\frac{15}{2}}{y}^6$

$=126x^{\frac{15}{2}}{y}^6$.