# Find the constant term in this binomial expansion?

## ${\left(2 {x}^{2} - \frac{1}{x}\right)}^{6}$ FInd the constant term in the expansion of this binomial.

Apr 2, 2017

$60.$

#### Explanation:

The General Term, denoted by, ${T}_{r + 1} ,$ in the Expansion of

${\left(a + b\right)}^{n}$ is,  T_(r+1)=""_nC_ra^(n-r)b^r, r=0,1,,...,n.

With, a=2x^2, b=-1/x, n=6, T_(r+1)=""_6C_r(2x^2)^(6-r)(-1/x)^r.

=""_6C_r(2)^(6-r)(-1)^r(x)^(12-2r)x^(-r)

=""_6C_r(2)^(6-r)(-1)^rx^(12-3r)..............(ast)

For the Const. Term, the index of $x$ must be $0$.

$\therefore 12 - 3 r = 0 \Rightarrow r = 4.$

(ast) rArr T_(4+1)=T_5=""_6C_4(2)^(6-4)(-1)^4x^(12-(3)(4)),

=""_6C_2*2^2*(1),

$= \frac{\left(6\right) \left(5\right)}{\left(1\right) \left(2\right)} \cdot 4.$

Hence, the desired const. term is $60$, and is the ${5}^{t h}$ term in

the Expansion.

Enjoy Maths.!