A)
Using the Binomial Theorem:
[tex](a+b)^n=\sum ^n_{k\mathop=0}\begin{pmatrix}n \\ k\end{pmatrix}a^{n-k}b^k[/tex]Then, replace for a=3x⁵, b=-1/9 y³ and n=4 to find the summation notation that Harold uses to express the expansion:
[tex]\mleft(3x^5-\frac{1}{9}y^3\mright)^4=\displaystyle\sum ^4_{k=0}\begin{pmatrix}4 \\ k\end{pmatrix}\mleft(3x^5\mright)^{4-k}\mleft(-\frac{1}{9}y^3\mright)^k[/tex]B)
To find the simplified terms of the expansion, expand the sum using the corresponding values of k from 0 to 4:
[tex]\begin{aligned}\mleft(3x^5-\frac{1}{9}y^3\mright)^4\quad = & \quad \begin{pmatrix}4 \\ 0\end{pmatrix}(3x^5)^{4-0}(-\frac{1}{9}y^3)^0 \\ & +\begin{pmatrix}4 \\ 1\end{pmatrix}(3x^5)^{4-1}(-\frac{1}{9}y^3)^1 \\ & +\begin{pmatrix}4 \\ 2\end{pmatrix}(3x^5)^{4-2}(-\frac{1}{9}y^3)^2 \\ & +\begin{pmatrix}4 \\ 3\end{pmatrix}(3x^5)^{4-3}(-\frac{1}{9}y^3)^3 \\ & +\begin{pmatrix}4 \\ 4\end{pmatrix}(3x^5)^{4-4}(-\frac{1}{9}y^3)^4 \\ \end{aligned}[/tex]Simplify the exponents of the parentheses (3x^5):
[tex]\begin{aligned}\mleft(3x^5-\frac{1}{9}y^3\mright)^4\quad = & \quad \begin{pmatrix}4 \\ 0\end{pmatrix}(3x^5)^4(-\frac{1}{9}y^3)^0 \\ & +\begin{pmatrix}4 \\ 1\end{pmatrix}(3x^5)^3(-\frac{1}{9}y^3)^1 \\ & +\begin{pmatrix}4 \\ 2\end{pmatrix}(3x^5)^2(-\frac{1}{9}y^3)^2 \\ & +\begin{pmatrix}4 \\ 3\end{pmatrix}(3x^5)^1(-\frac{1}{9}y^3)^3 \\ & +\begin{pmatrix}4 \\ 4\end{pmatrix}(3x^5)^0(-\frac{1}{9}y^3)^4 \\ \end{aligned}[/tex]Simplify the binomial coefficients:
[tex]\begin{aligned}\mleft(3x^5-\frac{1}{9}y^3\mright)^4\quad = & \quad 1\cdot(3x^5)^4(-\frac{1}{9}y^3)^0 \\ & +4\cdot(3x^5)^3(-\frac{1}{9}y^3)^1 \\ & +6\cdot(3x^5)^2(-\frac{1}{9}y^3)^2 \\ & +4\cdot(3x^5)^1(-\frac{1}{9}y^3)^3 \\ & +1\cdot(3x^5)^0(-\frac{1}{9}y^3)^4 \\ \end{aligned}[/tex]Elevate each parenthesis to its corresponding power:
[tex]\begin{aligned}\mleft(3x^5-\frac{1}{9}y^3\mright)^4\quad = & \quad (81x^{20})(1) \\ & +4(27x^{15})(-\frac{1}{9}y^3) \\ & +6(9x^{10})(\frac{1}{81}y^6) \\ & +4(3x^5)(-\frac{1}{729}y^9) \\ & +(1)(\frac{1}{6561}y^{12}) \\ \end{aligned}[/tex]Next, multiply all the factors on each term:
[tex]\begin{aligned}\mleft(3x^5-\frac{1}{9}y^3\mright)^4\quad = & \quad 81x^{20} \\ & -\frac{108}{9}x^{15}y^3 \\ & +\frac{54}{81}x^{10}y^6 \\ & -\frac{12}{729}x^5y^9 \\ & +\frac{1}{6561}y^{12} \\ \end{aligned}[/tex]Finally, simplify all the coefficients:
[tex]\begin{aligned}\mleft(3x^5-\frac{1}{9}y^3\mright)^4\quad = & \quad 81x^{20} \\ & -12x^{15}y^3 \\ & +\frac{2}{3}x^{10}y^6 \\ & -\frac{4}{243}x^5y^9 \\ & +\frac{1}{6561}y^{12} \\ \end{aligned}[/tex]Therefore, the simplified terms of the expansion are:
[tex]\mleft(3x^5-\frac{1}{9}y^3\mright)^4\; =\; 81x^{20}-12x^{15}y^3+\frac{2}{3}x^{10}y^6-\frac{4}{243}x^5y^9+\frac{1}{6561}y^{12}[/tex]
