Question

I need some help with this, it’s from my trigonometry prep guide. It asks to answer (a) and (b) Put these separately so I know which is which ^

Answer 1

Given the binomial below

$(3x^5-\frac{1}{9}y^3)^4$

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

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

Substitute for a and b into the formula above

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

Hence, the answer is

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

b) For the simplified terms of the expansion,

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

Hence, the answer is

$81x^{20}-12x^{15}y^3+\frac{2x^{10}y^6}{3}-\frac{4x^5y^9}{243}+\frac{y^{12}}{6561}$