Question

Let the term independent of $x$ in the expansion of ${\left(x^2-\frac{1}{x}\right)}^9$ has the value $p$ and $q$ be the sum of the coefficients of its middle terms, then $(p-q)$ equals

• A. $0$
• B. $2\cdot {}^9{C}_4$
• C. ${}^9{C}_5$
• D. ${}^9{C}_3$

Answer 1

Answer: (D)

$T_{r+1}$ in ${\left(x^2-\frac{1}{x}\right)}^9$ is ${}^9{C}_r{x}^{2(9-r)}{\left(-\frac{1}{x}\right)}^r$
$={}^9{C}_r\cdot x^{18-3r}\cdot (-1{)}^r$
for term independent of $x,$
$18-3r=0$
$\Rightarrow r=6$
$\therefore 7^{\text{th }}$ term is independent of $x$ and equals ${}^9{C}_6={}^9{C}_3=84$
Also there are $10$ terms, hence $5^{\text{th }}$ term and $6^{\text{th }}$ are the two middle term
$T_5={}^9{C}_4\cdot x^6$
$T_6=-{}^9{C}_5\cdot x^3$
$\therefore q=$ coefficient of $5^{\text{th }}+$ coefficient of $6^{\text{th }}$ term
Hence $p=84;q=0$
$\therefore p-q={}^9{C}_3$