Solution:
Given:
[tex](3x^5-\frac{1}{9}y^3)^4[/tex]Using binomial theorem expansion, the sum in summation notation is;
[tex](x+y)^n=\sum ^n_{k\mathop=0}\begin{bmatrix}{n} & {} \\ {k} & \end{bmatrix}x^{n-k}y^k[/tex]Comparing this to the expression given,
[tex]\begin{gathered} x=3x^5 \\ y=-\frac{1}{9}y^3 \end{gathered}[/tex]Question a:
The summation notation is;
[tex]\begin{gathered} (x+y)^n=\sum ^n_{k\mathop{=}0}\begin{bmatrix}{n} & {} \\ {k} & \end{bmatrix}x^{n-k}y^k \\ \\ (3x^5-\frac{1}{9}y^3)^4=\sum ^4_{k\mathop{=}0}\begin{bmatrix}{4} & {} \\ {k} & \end{bmatrix}(3x^5)^{4-k}(-\frac{1}{9}y^3)^k \end{gathered}[/tex]Question b:
The expansion can be as shown below;
[tex]\begin{gathered} (3x^5-\frac{1}{9}y^3)^4=\sum ^4_{k\mathop{=}0}\begin{bmatrix}{4} & {} \\ {k} & \end{bmatrix}(3x^5)^{4-k}(-\frac{1}{9}y^3)^k \\ =^4C_0(3x^5)^4+^4C_1(3x^5)^3(-\frac{1}{9}y^3)+^4C_2(3x^5)^2(-\frac{1}{9}y^3)^2+^4C_3(3x^5)^{}(-\frac{1}{9}y^3)^3+^4C_4(-\frac{1}{9}y^3)^4 \\ =1(3x^5)^4+4(3x^5)^3(-\frac{1}{9}y^3)+6(3x^5)^2(-\frac{1}{9}y^3)^2+4(3x^5)^{}(-\frac{1}{9}y^3)^3+1(-\frac{1}{9}y^3)^4 \\ =81x^{20}-12x^{15}y^3+\frac{2x^{10}y^6}{3}-\frac{4x^5y^9}{243}+\frac{y^{12}}{6561} \end{gathered}[/tex]Therefore, the simplified terms of the expansion is;
[tex]81x^{20}-12x^{15}y^3+\frac{2x^{10}y^6}{3}-\frac{4x^5y^9}{243}+\frac{y^{12}}{6561}[/tex]