Find the constant term in this binomial expansion?

(2x^2-1/x)^6
FInd the constant term in the expansion of this binomial.

1 Answer
Apr 2, 2017

60.

Explanation:

The General Term, denoted by, T_(r+1), in the Expansion of

(a+b)^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.

:. 12-3r=0 rArr r=4.

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

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

=((6)(5))/((1)(2))*4.

Hence, the desired const. term is 60, and is the 5^(th) term in

the Expansion.

Enjoy Maths.!