Question

The coefficient of $x^{-15}$ in $(3x^2-\frac{1}{3x^3}{)}^{10}$

• A. $\frac{40}{27}$
• B. $-\frac{40}{27}$
• C. $\frac{40}{81}$
• D. $-\frac{40}{81}$

Answer 1

Answer: (B)

$-\frac{40}{27}$

$T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$
Applying to the above question we get

$T_{r+1}=(-1){}^r{}^{10}{C}_r{x}^{20-2r}{3}^{10-r}{3}^{-r}{x}^{-3r}$

$=(-1){}^r{}^{10}{C}_r{x}^{20-5r}{3}^{10-3r}$ ...(i)

For the coefficient of $x^{-15}$

$20-5r=-15$

$35=5r$

$r=7$

Substituting in (i) we get
$T_8=-{}^{10}{C}_7{x}^{-15}{3}^{-4}$

$=-{}^{10}{C}_3{x}^{-15}{3}^{-4}$

$=-\frac{\mathrm{10.9.8}}{3!3^4}$

$=\frac{-40}{27}$