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

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Answer 1 · 2017-02-16T06:49:06

$T_{9} = 32476950000 x^{4} y^{8} .$ $T_9=32476950000x^4y^8.$

In the Expansion of Binomial ${(a + b)}^{n},$ $(a+b)^n,$ the General ${(r + 1)}^{t h}$ $(r+1)^(th)$ 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.!

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