Question

$5$th term from the end in the expansion of ${\left[\left(\frac{x^3}{2}\right)-\left(\frac{2}{x^2}\right)\right]}^{12}$ is

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

Answer 1

The correct option is C

$7920x^{-4}$

Explanation for the correct option:

Step 1: Find the position of the required term from the beginning:

Subtract $5$ from total terms of the expansion to get the position from the beginning.

Term from beginning $=$power of expression$+1-$position of term from end

$\begin{array}{rcl}n& =& 12+1-5\\ & =& 9\end{array}$

Step 2: Find the $9$th term of the given expression.

Use the binomial expansion formula to find $9$th term.

$$\begin{gathered} T_9={}^{12}C_{9-1}{\left(\frac{x^3}{2}\right)}^{12-9+1}{\left(-\frac{2}{x^2}\right)}^{9-1} \\ ={}^{12}C_8{\left(\frac{x^3}{2}\right)}^4{\left(-\frac{2}{x^2}\right)}^8 \\ =7920x^{-4} \end{gathered}$$ $\left[\because T_r={}^{n}C_{r-1}{a}^{n-r+1}{b}^{r-1}\right]$

Hence, option (C) is the correct answer.