Question

The 4th term from end in the expansion of ${(\frac{3}{x^2}-\frac{x^3}{5})}^8$.

• A. $\frac{-56x^6}{5^4}$
• B. $\frac{-9x^9}{5^5}$
• C. $\frac{-56x^{1}5}{5^{1}55}$
• D. None of these

Answer 1

The correct option is D None of these

the power given is 8 hence there would be 9 terms

$4^{th}$ term from end means $6^{th}$ term from begining

$T_6$=$T_{5+1}$=${}^8{C}_5$.${\left(\frac{3}{x^2}\right)}^{8-5}$.${\left(\frac{-x^3}{5}\right)}^5$

$T_6$=${}^8{C}_5$.$3^3$.$\frac{-1}{5^5}$.$x^9$

$T_6$=$-56$$\frac{{\left(3\right)}^3}{{\left(5\right)}^5}$.$x^9$