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. $7220\,x^{-4}$
• D. $-7520\,x^4$

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\times11\times 10\times 9}{1\times 2\times 3\times 4}\cdot \frac{2^{8}1}{2^{4}\,x^{4}}$
$= \frac{12\times 11\times 10\times 9}{24}\times \frac{16}{x^{4}}$
$= \frac{11\times 5\times 9\times 16}{x^{4}} = \frac{99\times 80}{x^{4}}$
$= \frac{7920}{x^{4}}$