Find the constant term in this binomial expansion?

Question

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

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Answer 1 · 2017-04-02T10:48:32

#60.#

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.!