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

1 Answer
Aug 3, 2017

Answer:

# a=45.#

Explanation:

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.#