The constant term in the bionomial expansion of $$\left(3x^2 - \frac{1}{4x}\right)^6$$ is

Solution:

The constant term in the bionomial expansion of $$\left(3x^2 - \frac{1}{4x}\right)^6$$ is

$$^{n}C_r$$  $$\left(3x^2\right)^{\left(6-r\right)}\left(-\frac{1}{4x}\right)^r$$

$$3^{\left(6-r\right)}x^{\left(12-2r\right)}\frac{\left(-1\right)^r}{4^rx^r}$$

= Constant $$x^{\left(12-2r\right)}x^(-r)$$

= Constant$$x^{\left(12-2r\ -r\right)}\ =Constant\ x^{\left(12-3r\right)}$$

For constant term , power of x should be 0

12-3r = 0

r =4

Constant = $$^6Cr\ 3^{\left(6-r\right)}\ \frac{\left(-1\right)^r}{\left(4\right)^r}\ =\ \frac{6!}{\left(6-2\right)!2!}.3^{\left(6-4\right)}\left(-\frac{1}{4}\right)^4$$

=$$=\ \frac{6!}{\left(4\right)!2!}.3^{\left(2\right)}\frac{1}{4^4}^{ }=\frac{5.6}{2}.9.\left(\frac{1}{256}\right)$$

= 135/256