Question

I need help with this Struggling It asks to answer (a) and (b) Put these separately ^ so I know which is which

Answer 1

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=n)_(r\mathop=0)\begin{bmatrix}{n} \\ {r}\end{bmatrix}a^(n-r)\cdot b^r`

Using the formula and substitute

`a=3x^5,b=-(1)/(9)y^3\text{ and n=4}`

Therefore,

`(3x^5-(1)/(9)y^3)^4=\sum ^(r=4)_(r\mathop=0)\begin{bmatrix}{4} \\ {r}\end{bmatrix}(3x^5)^(4-r)\cdot(-(1)/(9)y^3)^r`

Hence, the sum in summation notation that he uses to express the expansion is

`(3x^5-(1)/(9)y^3)^4=\sum ^(r=4)_{r\mathop{=}0}\begin{bmatrix}{4} \\ {r}\end{bmatrix}(3x^5)^(4-r)\cdot(-(1)/(9)y^3)^r`

b) Let us now write the simplified terms of the expansion.

Therefore,

Using Combination formula to expand the expression above

The combination formula is,

`^nC_r=(n!)/(r!(n-r)!)`

Where,

`n!\text{ = n(n-1)(n-2)}\ldots3.2.1`

Hence,

`\begin{gathered} =(4!)/(0!\left(4-0\right)!)\mleft(3x^5\mright)^4\mleft(-(1)/(9)y^3\mright)^0+(4!)/(1!\left(4-1\right)!)\mleft(3x^5\mright)^3\mleft(-(1)/(9)y^3\mright)^1+(4!)/(2!\left(4-2\right)!)\mleft(3x^5\mright)^2\mleft(-(1)/(9)y^3\mright)^2 \\ +(4!)/(3!\left(4-3\right)!)\mleft(3x^5\mright)^1\mleft(-(1)/(9)y^3\mright)^3+(4!)/(4!\left(4-4\right)!)\mleft(3x^5\mright)^0\mleft(-(1)/(9)y^3\mright)^4 \end{gathered}`

Simplifying the above, we have

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

Hence, the simplified terms of the expansion are

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