How do you find the 9th term in the expansion of the binomial #(10x-3y)^12#?

1 Answer
Feb 16, 2017

Answer:

#T_9=32476950000x^4y^8.#

Explanation:

In the Expansion of Binomial #(a+b)^n,# the General #(r+1)^(th)# Term

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

In our Problem,

#a=10x, b=-3y, n=12, and, r+1=9, i.e., r=8.#

#:. T_9=""_12C_8(10x)^(12-8)(-3y)^8#

#=""_12C_4(10)^4(3)^8x^4y^8,#

#={(12)(11)(10)(9)}/{(1)(2)(3)(4)}(10000)(6561)x^4y^8#

#=(11)(5)(9)(10000)(6561)x^4y^8#

#T_9=32476950000x^4y^8#

Enjoy Maths.!