Given the expression:

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:

Therefore, the sum in summation nottation will be:

• (b). Write the simplified terms of the expansion.
To write the simplified terms, we have:
![\begin{gathered} (4!)/(4-0)!0!)*(3x^5)^(4-0)\cdot(-(1)/(9)y^3)^0+(4!)/((4-1)!1!)\cdot(3x^5)^(4-1)\cdot(-(1)/(9)y^3)^1+(4!)/((4-2)!2!)\cdot(3x^5)^(4-2)\cdot(-(1)/(9)y^3)^2+(4!)/((4-3)!3!)\cdot(3x^5)^(4-3)\cdot(-(1)/(9)y^3)^3+(4!)/((4-4)!4!)\cdot(3x^5)^(4-4)\cdot(-(1)/(9)y^3)^4 \\ \\ \\ 81x^(20)-12x^(15)y^3+(2x^(10)y^6)/(3)-(4x^5y^9)/(243)+(y^(12))/(6561) \end{gathered}]()
ANSWER:
