Given the expression:
[tex](3x^5-\frac{1}{9}y^3)^4[/tex]Given that Harold uses boinomial theorem to expand the given binomial, let's solve for the following:
• (a). What is the sum in summation notation that he uses the express the expnansion.
Apply the formula:
[tex]\sum ^n_{k=0}nCk(a^{n-k}b^k)[/tex]Therefore, the sum in summation nottation will be:
[tex]\sum ^4_{k=0}\frac{4!}{(4-k)!k!}\cdot(3x^5)^{4-k}\cdot(-\frac{1}{9}y^3)^k[/tex]• (b). Write the simplified terms of the expansion.
To write the simplified terms, we have:
[tex]\begin{gathered} \frac{4!}{4-0)!0!}\times(3x^5)^{4-0}\cdot(-\frac{1}{9}y^3)^0+\frac{4!}{(4-1)!1!}\cdot(3x^5)^{4-1}\cdot(-\frac{1}{9}y^3)^1+\frac{4!}{(4-2)!2!}\cdot(3x^5)^{4-2}\cdot(-\frac{1}{9}y^3)^2+\frac{4!}{(4-3)!3!}\cdot(3x^5)^{4-3}\cdot(-\frac{1}{9}y^3)^3+\frac{4!}{(4-4)!4!}\cdot(3x^5)^{4-4}\cdot(-\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]ANSWER:
[tex]\begin{gathered} (a)\text{. }\sum ^4_{k=0}\frac{4!}{(4-k)!k!}\cdot(3x^5)^{4-k}\cdot(-\frac{1}{9}y^3)^k \\ \\ \\ (b)\text{. }81x^{20}-12x^{15}y^3+\frac{2x^{10}y^6}{3}-\frac{4x^5y^9}{243}+\frac{y^{12}}{6561} \end{gathered}[/tex]