Question

Find the $7$th term in the expansion of ${(\frac{4x}{5}-\frac{5}{2x})}^9$.

Answer 1

The binominal expansion of ${(\frac{4x}{5}-\frac{5}{2x})}^9$

$={}^a{C}_0{\left(\frac{-5}{2x}\right)}^0{\left(\frac{4x}{5}\right)}^{9-0}{+}^9{C}_1{\left(\frac{-5}{2x}\right)}^1{\left(\frac{4x}{5}\right)}^{9-1}+.....{+}^a{C}_9{\left(\frac{-5}{2x}\right)}^9{\left(\frac{4x}{5}\right)}^{9-9}$

$=\sum _{r=0}^9{}^a{C}_r{\left(\frac{-5}{2x}\right)}^r{\left(\frac{4x}{5}\right)}^{9-r}$

the $7^{th}$ term would br for $r=6$

i.e, ${}^9{C}_6{\left(\frac{-5}{2x}\right)}^6{\left(\frac{4x}{5}\right)}^3$

$=84.\frac{5^6}{2^6.x^6}.\frac{4^3.x^3}{5^3}$

$={84.5}^3{x}^{-3}$