Part (a).
Sigma notation (or summation notation) of binomial expansion is the following:
[tex](w+z)^n=\sum ^n_{k\mathop=0}\binom{n}{k}w^{n-k}\cdot z^k[/tex]where
[tex]\binom{n}{k}[/tex]denotes the binomial coefficient.
In our case, n is 4 and
[tex]\begin{gathered} w=3x^5 \\ z=-\frac{1}{9}y^3 \end{gathered}[/tex]So by substituting these terms into the sigma expantion, we have
[tex](3x^5+(-\frac{1}{9}y^3))^4=\sum ^4_{k\mathop{=}0}\binom{4}{k}(3x^5)^{4-k}\cdot(-\frac{1}{9}y^3)^k[/tex]So, the sum in summation notation is:
[tex](3x^5-\frac{1}{9}y^3)^4=\sum ^4_{k\mathop{=}0}\binom{4}{k}(3x^5)^{4-k}\cdot(-\frac{1}{9}y^3)^k[/tex]Part b.
By expanding the above sum, we have
[tex]\begin{gathered} (3x^5-\frac{1}{9}y^3)^4=\binom{4}{0}(3x^5)^4\cdot(-\frac{1}{9}y^3)^0+\binom{4}{1}(3x^5)^3\cdot(-\frac{1}{9}y^3)^1+\binom{4}{2}(3x^5)^2\cdot(-\frac{1}{9}y^3)^2+ \\ \binom{4}{3}(3x^5)^1\cdot(-\frac{1}{9}y^3)^3+\binom{4}{4}(3x^5)^0\cdot(-\frac{1}{9}y^3)^4 \\ \end{gathered}[/tex]Since
[tex]\begin{gathered} \binom{4}{0}=1 \\ \binom{4}{1}=4 \\ \binom{4}{2}=6 \\ \binom{4}{3}=4 \\ \binom{4}{4}=1 \end{gathered}[/tex]we have
[tex](3x^5-\frac{1}{9}y^3)^4=(3x^5)^4+4(3x^5)^3\cdot(-\frac{1}{9}y^3)^{}+6(3x^5)^2\cdot(-\frac{1}{9}y^3)^2+4(3x^5)^1\cdot(-\frac{1}{9}y^3)^3+(-\frac{1}{9}y^3)^4[/tex]which gives
[tex](3x^5-\frac{1}{9}y^3)^4=81x^{20}-12x^{15}\cdot y^3+\frac{6}{9}x^{10}\cdot y^6-\frac{12}{729}x^5\cdot y^9+\frac{1}{6561}y^{12}[/tex]Therefore, the simplified expansion is given by:
[tex](3x^5-\frac{1}{9}y^3)^4=81x^{20}-12x^{15}\cdot y^3+\frac{2}{3}x^{10}\cdot y^6-\frac{4}{243}x^5\cdot y^9+\frac{1}{6561}y^{12}[/tex]