Question

The $8^{\text{th}}$ term of ${(3x+\frac{2}{3x^2})}^{12}$ when expanded in ascending powers of x, is $\frac{a}{x^n},$ where a & n are positive integers, Find the value of $(a-n)$

Answer 1

Given

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

8th term ${}^{12}{C}_7(3x{)}^{12-7}{\left(\frac{2}{3x^2}\right)}^7$

$\frac{a}{x^n}{=}^{12}{C}_7(3x{)}^5\frac{2^7}{(3x^2{)}^7}$

$\frac{a}{x^n}{=}^{12}{C}_7{3}^5{x}^5\frac{2^7}{3^7{x}^{14}}$

$\frac{a}{x^n}{=}^{12}{C}_7{3}^{5-7}{2}^7{x}^{5-14}$

$\frac{a}{x^n}{=}^{12}{C}_7{3}^{-2}{2}^7{x}^{-9}$

$\frac{a}{x^n}=\frac{{}^{12}{C}_7{3}^{-2}{2}^7}{x^9}$

$\frac{a}{x^n}=\frac{11264}{x^9}$

$\Rightarrow a=11264$

$n=9$

$a-n=11264-9$

$a-n=11255$