Question

Which term is the constant term in the expansion of ${(2x-\frac{1}{3x})}^6$?

• A. $2nd$ term
• B. $3rd$ term
• C. $4th$ term
• D. $5th$ term

Answer 1

The correct option is C $4th$ term

Given,

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

General term

$T_{r+1}={}^n{C}_r{x}^{n-r}{y}^r$

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

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

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

For constant term

$6-2r=0$

$6=2r$

$r=3$

Hence, $T_{r+1}=T_{3+1}=T_4$ will be constant term

Hence correct option is C.