How do you find the coefficient of #a# of the term #az^4t^8# in the expansion of the binomial #(z^2-t)^10#?

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Answer 1 · 2017-08-03T11:30:48

# a=45.#

We know that, the General #(r+1)^(th)# Term, denoted by,

#T_(r+1),# in the Expansion of #(a+b)^n,# is given by,

#T_(r+1)=""_nC_ra^(n-r)b^r, r=0,1,2,...,n.#

In our Problem, #a=z^2, b=-t, n=10.#

#:. T_(r+1)=""_10C_r(z^2)^(10-r)(-t)^r,#

#=""_10C_r*z^(20-2r)(-1)^r*t^r, r=0,1,2,...,10.#

Comparing this with, #az^4t^8, # we have,

#a=(-1)^r""_10C_r, 20-2r=4, and, r=8.#

#:. a=(-1)^8""_10C_8=""_10C_2=(10*9)/(1*2)=45.#