Question

The fifth term from the end in the expansion ${\left(\frac{x^3}{2}-\frac{2}{x^2}\right)}^{12}$ is

• A. $\frac{7920}{x^4}$
• B. $-\frac{7920}{x^4}$
• C. $7220x^{-4}$
• D. $-7520x^4$

Answer 1

Answer: (D)

Fifth term, from the end has $8$ terms before it $=T_{8+1}$
$={}^{12}{c}_8{\left(\frac{x^3}{2}\right)}^4{\left(-\frac{2}{x^2}\right)}^8$
$=\frac{12\times 11\times 10\times 9}{1\times 2\times 3\times 4}\cdot \frac{2^{8}1}{2^4{x}^4}$
$=\frac{12\times 11\times 10\times 9}{24}\times \frac{16}{x^4}$
$=\frac{11\times 5\times 9\times 16}{x^4}=\frac{99\times 80}{x^4}$
$=\frac{7920}{x^4}$