Question

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

Answer 1

Part (a).

Sigma notation (or summation notation) of binomial expansion is the following:

`(w+z)^n=\sum ^n_(k\mathop=0)\binom{n}{k}w^(n-k)\cdot z^k`

where

`\binom{n}{k}`

denotes the binomial coefficient.

In our case, n is 4 and

`\begin{gathered} w=3x^5 \\ z=-(1)/(9)y^3 \end{gathered}`

So by substituting these terms into the sigma expantion, we have

`(3x^5+(-(1)/(9)y^3))^4=\sum ^4_{k\mathop{=}0}\binom{4}{k}(3x^5)^(4-k)\cdot(-(1)/(9)y^3)^k`

So, the sum in summation notation is:

`(3x^5-(1)/(9)y^3)^4=\sum ^4_{k\mathop{=}0}\binom{4}{k}(3x^5)^(4-k)\cdot(-(1)/(9)y^3)^k`

Part b.

By expanding the above sum, we have

`\begin{gathered} (3x^5-(1)/(9)y^3)^4=\binom{4}{0}(3x^5)^4\cdot(-(1)/(9)y^3)^0+\binom{4}{1}(3x^5)^3\cdot(-(1)/(9)y^3)^1+\binom{4}{2}(3x^5)^2\cdot(-(1)/(9)y^3)^2+ \\ \binom{4}{3}(3x^5)^1\cdot(-(1)/(9)y^3)^3+\binom{4}{4}(3x^5)^0\cdot(-(1)/(9)y^3)^4 \\ \end{gathered}`

Since

`\begin{gathered} \binom{4}{0}=1 \\ \binom{4}{1}=4 \\ \binom{4}{2}=6 \\ \binom{4}{3}=4 \\ \binom{4}{4}=1 \end{gathered}`

we have

`(3x^5-(1)/(9)y^3)^4=(3x^5)^4+4(3x^5)^3\cdot(-(1)/(9)y^3)^{}+6(3x^5)^2\cdot(-(1)/(9)y^3)^2+4(3x^5)^1\cdot(-(1)/(9)y^3)^3+(-(1)/(9)y^3)^4`

which gives

`(3x^5-(1)/(9)y^3)^4=81x^(20)-12x^(15)\cdot y^3+(6)/(9)x^(10)\cdot y^6-(12)/(729)x^5\cdot y^9+(1)/(6561)y^(12)`

Therefore, the simplified expansion is given by:

`(3x^5-(1)/(9)y^3)^4=81x^(20)-12x^(15)\cdot y^3+(2)/(3)x^(10)\cdot y^6-(4)/(243)x^5\cdot y^9+(1)/(6561)y^(12)`