Question

The $8^{th}$ term of ${(3x+\frac{2}{3x^2})}^{12}$, when expanded ina scending power of $x$, is

• A. $\frac{228096}{x^3}$
• B. $\frac{228096}{x^9}$
• C. $\frac{328179}{x^3}$
• D. None of these

Answer 1

Answer: (A)

$\frac{228096}{x^3}$

When ${(3x+\frac{2}{3x^2})}^{12}$ is expanded, the power of $x$ goes on decreasing as the terms proceed.

hence, it is expanded in descending power of $x$.

So ${(\frac{2}{3x^2}+3x)}^{12}$, when expanded will be ascending power of $x$.

Now $t_8$ in ${(\frac{2}{3x^2}+3x)}^{12}$ ${=}^{12}{C}_7{\left(\frac{2}{3x^2}\right)}^{12-7}.{\left(3x\right)}^7$

$=\frac{12!}{7!5!}.{\left(\frac{2}{3x^2}\right)}^5.{\left(3x\right)}^7=\frac{12\times 11\times 10\times 9\times 8}{5\times 4\times 3\times 2}.\frac{2^5.3^2}{x^3}$ $=\frac{228096}{x^3}$