Answer:
The sum in summation notation is:
[tex]\sum ^4_{r\mathop{=}0}(-1)^r(4Crx^{4-r}y^r)[/tex]
The expansion is:
[tex]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]
Explanation:
Given the expression:
[tex](3x^5-\frac{1}{9}x^3)^4[/tex]
In summation notation, this can be written as:
[tex]\sum ^4_{r\mathop=0}(-1)^r(4Crx^{4-r}y^r)[/tex]
The simplified terms of the expression is:
[tex]\begin{gathered} 4C0(3x^5)^{\mleft\{4-0\mright\}}(\frac{1}{9}y^3)^0-4C1(3x^5)^{\mleft\{4-1\mright\}}(\frac{1}{9}y^3)^1+4C2(3x^5)^{\mleft\{4-2\mright\}}(\frac{1}{9}y^3)^2-4C3(3x^5)^{\mleft\{4-3\mright\}}(\frac{1}{9}y^3)^3+4C4(3x^5)^{\mleft\{4-4\mright\}}(\frac{1}{9}y^3)^4 \\ \\ =(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+(\frac{1}{9}y^3)^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} \end{gathered}[/tex]
