Question

Find the middle term in the expansion of ${(\frac{{2x}^2}{3}-\frac{3}{2x})}^{12}$.

Answer 1

$(\frac{2x^2}{3}-\frac{3}{2x}{)}^{12}$

middle term be r=7

${}^{12}{C}_7(\frac{2x^2}{3}{)}^5.(\frac{-3}{2x}{)}^7$

${-}^{12}{C}_7{2}^5.\frac{x^{10}}{3^5}\frac{3^7}{2^7.x^7}$

${-}^{12}{C}_7\frac{3^2}{2^2}.x^3$

$-\frac{12!}{5!7!}\frac{9}{4}.x^3$

$\frac{12\times 11\times 10\times 9\times 8\times 9}{5\times 4\times 3\times 2\times 4}.x^3$

$-1782x^3$