Question

The term independent of $x$ in the expansion of ${(2x+\frac{1}{3x})}^6$ is

• A. $\frac{160}{9}$
• B. $\frac{80}{9}$
• C. $\frac{160}{27}$
• D. $\frac{80}{3}$

Answer 1

The correct option is C $\frac{160}{27}$

Given,

${(2x+\frac{1}{3x})}^6$

$T_{r+1}{=}^n{C}_r\left(a{)}^{n-r}\right(b{)}^r$

$n=6,a=2x,b=\frac{1}{3x}$

$T_{r+1}{=}^6{C}_r(2x{)}^{6-r}{\left(\frac{1}{3x}\right)}^r$

$T_{r+1}{=}^6{C}_r{2}^{6-r}{\left(\frac{1}{3}\right)}^r{x}^{6-2r}$

equate x to the power to zero

$x^{6-2r}=x^0$

$6-2r=0$

$\Rightarrow r=3$

substitute the value of r back to the equation

$T_{3+1}{=}^6{C}_3{2}^{6-3}{\left(\frac{1}{3}\right)}^3{x}^{6-2\left(3\right)}$

$T_4=\frac{6!}{3!(6-3)!}\times \frac{8}{27}\times x^0$

$T_4=\frac{720}{6\times 6}\times \frac{8}{27}$

$\therefore T_4=\frac{160}{27}$