Question

The coefficient of $x^3$ in ${(\sqrt{x^5}+\frac{3}{\sqrt{x^3}})}^6$ is

• A. $0$
• B. $120$
• C. $420$
• D. $540$

Answer 1

Answer: (D)

$540$

$T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$
Applying to the above question we get
$T_{r+1}={}^6{C}_r\left(x^{15-\frac{5r}{2}}\right)\left(3^r{x}^{\frac{-3r}{2}}\right)$
$={}^6{C}_r\left(x^{15-4r}\right)\left(3^r\right)$ ...(i)
For the coefficient of $x^3$
$15-4r=3$
$4r=12$
$r=3$
Substituting in equation (i), we get
$T_4={}^6{C}_3\left(3\right){}^3\left(x\right){}^3$
Coefficient $=\frac{\mathrm{6.5.4.}\left(3^3\right)}{3!}$
$=540$