Question

I need help with this practice problem It asks to answer (a) & (b) Please separate (a) & (b) so I know which one is which

Answer 1

Given:

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

Using binomial theorem expansion, the sum in summation notation is;

`(x+y)^n=\sum ^n_(k\mathop=0)\begin{bmatrix}{n} & {} \\ {k} & \end{bmatrix}x^(n-k)y^k`

Comparing this to the expression given,

`\begin{gathered} x=3x^5 \\ y=-(1)/(9)y^3 \end{gathered}`

Question a:

The summation notation is;

`\begin{gathered} (x+y)^n=\sum ^n_{k\mathop{=}0}\begin{bmatrix}{n} & {} \\ {k} & \end{bmatrix}x^(n-k)y^k \\ \\ (3x^5-(1)/(9)y^3)^4=\sum ^4_{k\mathop{=}0}\begin{bmatrix}{4} & {} \\ {k} & \end{bmatrix}(3x^5)^(4-k)(-(1)/(9)y^3)^k \end{gathered}`

Question b:

The expansion can be as shown below;

`\begin{gathered} (3x^5-(1)/(9)y^3)^4=\sum ^4_{k\mathop{=}0}\begin{bmatrix}{4} & {} \\ {k} & \end{bmatrix}(3x^5)^(4-k)(-(1)/(9)y^3)^k \\ =^4C_0(3x^5)^4+^4C_1(3x^5)^3(-(1)/(9)y^3)+^4C_2(3x^5)^2(-(1)/(9)y^3)^2+^4C_3(3x^5)^{}(-(1)/(9)y^3)^3+^4C_4(-(1)/(9)y^3)^4 \\ =1(3x^5)^4+4(3x^5)^3(-(1)/(9)y^3)+6(3x^5)^2(-(1)/(9)y^3)^2+4(3x^5)^{}(-(1)/(9)y^3)^3+1(-(1)/(9)y^3)^4 \\ =81x^(20)-12x^(15)y^3+(2x^(10)y^6)/(3)-(4x^5y^9)/(243)+(y^(12))/(6561) \end{gathered}`

Therefore, the simplified terms of the expansion is;

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