Question

I need help solving this, it’s from my trigonometry prep bookIt asks to answer (a) and (b)Please put these ^ separately so I know which is which

Answer 1

ANSWER

$\begin{gathered} (a)\sum ^4_{k\mathop{=}0}\begin{bmatrix}{} & 4 \\ {} & {k}\end{bmatrix}(3x^5)^{4-k}(-\frac{1}{9}y^3)^k \\ (b)81x^{20}-12x^{15}y^3+\frac{2}{3}x^{10}y^6-\frac{4}{243}x^5y^9+\frac{1}{6561}y^{12} \end{gathered}$

EXPLANATION

We want to expand the given expression using binomial expansion:

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

(a) In summation notation, binomial expansion is written as:

$(a+b)^n=\sum ^n_{k\mathop=0}\begin{bmatrix}{} & n \\ {} & {k}\end{bmatrix}a^{n-k}b^k$

where:

$\begin{gathered} a=3x^5 \\ b=-\frac{1}{9}y^3 \\ n=4 \end{gathered}$

Therefore, the summation notation for the expansion is:

$(3x^5-\frac{1}{9}y^3)^4=\sum ^4_{k\mathop{=}0}\begin{bmatrix}{} & 4 \\ {} & {k}\end{bmatrix}(3x^5)^{4-k}(-\frac{1}{9}y^3)^k$

(b) Now, we want to expand the expression. To do this, find each term taking k to go from 0 to 4, find the sum of the terms, and simplify.

That is:

$\begin{gathered} k=0\colon \\ ^{}\begin{bmatrix}{4} & {} \\ {0} & {}\end{bmatrix}(3x^5)^{4-0}(-\frac{y^3}{9})^0 \\ \frac{4!}{(4-0)!0!}(3x^5)^4 \\ \Rightarrow81x^{20} \end{gathered}$$\begin{gathered} k=1\colon \\ \begin{bmatrix}{4} & {} \\ {1} & {}\end{bmatrix}(3x^5)^{4-1}(\frac{-y^3}{9})^1_{} \\ \frac{4!}{(4-1)!1!}(3x^5)^3(\frac{-y^3}{9}) \\ \Rightarrow-12x^{15}y^3 \end{gathered}$$\begin{gathered} k=2\colon \\ \begin{bmatrix}{4} & {} \\ {2} & {}\end{bmatrix}(3x^5)^{4-2}(-\frac{y^3}{9})^2 \\ \frac{4!}{(4-2)!2!}(3x^5)^2(-\frac{y^3}{9})^2 \\ \Rightarrow\frac{2}{3}x^{10}y^6 \end{gathered}$$\begin{gathered} k=3\colon \\ \begin{bmatrix}{4} & {} \\ {3} & {}\end{bmatrix}(3x^5)^{4-3}(-\frac{y^3}{9})^3 \\ \frac{4!}{(4-3)!3!}(3x^5)(-\frac{y^3}{9})^3 \\ \Rightarrow-\frac{4}{243}x^5y^9 \end{gathered}$$\begin{gathered} k=4\colon \\ \begin{bmatrix}{4} & {} \\ {4} & {}\end{bmatrix}(3x^5)^{4-4}(-\frac{y^3}{9})^4 \\ \frac{4!}{(4-4)!4!}(-\frac{y^3}{9})^4^{} \\ \frac{1}{6561}y^{12} \end{gathered}$

Therefore, in simplified form, the expanded expression is:

$(3x^5-\frac{y^3}{9})^4=81x^{20}-12x^{15}y^3+\frac{2}{3}x^{10}y^6-\frac{4}{243}x^5y^9+\frac{1}{6561}y^{12}$