Question

Find the $7th$ term from the end in the expansion of $(2x^2-\frac{3}{2x}{)}^8$

Answer 1

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

There are $9$ terms in the expansion so, $7th$ term from back is $3^{rd}$ term from first

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

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

$=\frac{28\times 2^6}{4}\times 9\times x^{10}$

$=63\times 2^6\times x^{10}$

$=63\times 2^6{x}^{10}$

Hence, the answer is $63\times 2^6{x}^{10}.$