Question

$5^{th}$ terms from the end in the expansion of ${(\frac{x^3}{2}-\frac{2}{x^2})}^{12}$ is

• A. $-7920x^{-4}$
• B. $7920x^4$
• C. $7920x^{-4}$
• D. $-7920x^4$

Answer 1

The correct option is C $7920x^{-4}$

$T_{r+1}={}^n{C}_r\left(a{)}^{n-r}\right(b{)}^r$

Now,

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

$5^{th}$ term, from the end has $8$ before it $=T_{8+1}$

$T_{8+1}={}^{12}{C}_8{\left(\frac{x^3}{2}\right)}^4{(-\frac{2}{x^2})}^8$

$T_{8+1}=\frac{12\times 11\times 10\times 9}{1\times 2\times 3\times 4}\times \frac{2^8}{2^4{x}^4}$

$T_{8+1}=\frac{12\times 11\times 10\times 9}{24}\times \frac{16}{x^4}$

$T_{8+1}=\frac{11\times 5\times 9\times 16}{x^4}$

$T_{8+1}=\frac{7920}{x^4}=7920x^{-4}$