Question

Harold uses the binomial theorem to expand the binomial

(3x^5 -1/9y^3)^4

(a) What is the sum in summation notation that he uses to express the expansion?

(b) Write the simplified terms of the expansion.

Answer 1

Answer with Step-by-step explanation:

We are given that

`(3x^5-(1)/(9)y^3)^4`

a.We know that binomial theorem in summation form

`(a+b)^n=\sum_(r=0)^(r=n)\binom{n}{r}a^(n-r)\cdot b^r`

Using the formula and substitute
`a=3x^5,b=-(1)/(9)y^3` and n=4

`(3x^5-(1)/(9)y^3)^4=\sum_(r=0)^(r=4)\binom{4}{r}(3x^5)^(4-r)\cdot (-(1)/(9)y^3)^r`

b.
`(3x^5-(1)/(9)y^3)^4=\sum_(r=0)^(r=4)(3x^5)^(4-r)(-(1)/(9)y^3)^r=\binom{4}{0}(3x^5)^4+\binom{4}{1}(3x^5)^3(-(1)/(9)y^3)+\binom{4}{2}(3x^5)^2(-(1)/(9)y^3)^2+\binom{4}{3}(3x^5)(-(1)/(9)y^3)^3+\binom{4}{4}(-(1)/(9)y^3)^4`

Combination formula:
`nC_r=(n!)/(r!(n-r)!)`

`n!=n(n-1)(n-2)...3\cdot 2\cdot 1`

`(3x^5-(1)/(9)y^3)^4=(4!)/(0!4!)(81x^(20))+(4!)/(3!)(27x^(15))(-(1)/(9)y^3)+(4!)/(2!2!)(9x^(10))((1)/(81)y^6)+(4!)/(3!)(3x^5)(-(1)/(729)y^9)+(4!)/(4!)((1)/(6561)y^(12))`

`(3x^5-(1)/(9)y^3)^4=81x^(20)-(4* 3!)/(3!)(3x^(15)y^3+(4* 3* 2!)/(2* 1* 2!)* (1)/(9)x^(10)y^6-(4* 3!)/(3!)((x^5y^9)/(243)+\frac{6561}y^(12)`

`(3x^5-(1)/(9)y^3)^4=81x^(20)-12x^(15)y^3+(2)/(3)x^(10)y^6-(4)/(243)x^5y^9+(1)/(6561)y^(12)`