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Powlo · Answer 1 · 2023-12-18T02:29:09+0000

Given the binomial below

a) The formula for the summation notation of a binomial theorem is

$(a+b)^n=\sum ^n_(r=0)(^n_r)(a)^(n-r)(b)^r$

Where

$\begin{gathered} a=3x^5 \\ b=-(1)/(9)y^3 \end{gathered}$

Substitute for a and b into the formula above

$(3x^5-(1)/(9)y^3)^4=\sum ^4_(r=0)(^4_r)(3x^5)^(4-r)(-(1)/(9)y^3)^r$

Hence, the answer is

$\sum ^4_(r=0)(^4_r)(3x^5)^(4-r)(-(1)/(9)y^3)^r$

b) For the simplified terms of the expansion,

$\begin{gathered} \sum ^4_(r=0)(^4_r)(3x^5)^(4-r)(-(1)/(9)y^3)^r=(4!)/(0!\left(4-0\right)!)\mleft(3x^5\mright)^4\mleft(-(1)/(9)y^3\mright)^0+(4!)/(1!\left(4-1\right)!)\mleft(3x^5\mright)^3\mleft(-(1)/(9)y^3\mright)^1+(4!)/(2!\left(4-2\right)!)\mleft(3x^5\mright)^2\mleft(-(1)/(9)y^3\mright)^2+(4!)/(3!(4-3)!)(3x^5)^1(-(1)/(9)y^3)^3+(4!)/(4!(4-4)!)(3x^5)^0(-(1)/(9)y^3)^4 \\ =81x^(20)-12x^(15)y^3+(2x^(10)y^6)/(3)-(4x^5y^9)/(243)+(y^(12))/(6561) \end{gathered}$

Hence, the answer is

81x^(20)-12x^(15)y^3+(2x^(10)y^6)/(3)-(4x^5y^9)/(243)+(y^(12))/(6561)

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