# Hi, how do you answer just question (iii)? (given the context of question (i) and (ii))

## Question:

Suppose that {eq}\omega^3 = 1, \;\omega \neq 1 {/eq}, and k is a positive integer.

a) Find the two possible values of {eq}1 + \omega^k + \omega^{2k} {/eq}.

b) Use the binomial theorem to expand {eq}(1 + \omega)^n {/eq} and {eq}(1 + \omega^2)^n {/eq}, where n is a positive integer.

c) Let {eq}l {/eq} be the largest integer such that {eq}3l\leq n {/eq}.

Deduce that

{eq}\begin{pmatrix} n\\ 0 \end{pmatrix} + \begin{pmatrix} n\\ 3 \end{pmatrix} + \begin{pmatrix} n\\ 6 \end{pmatrix} + \cdots + \begin{pmatrix} n\\ 3l \end{pmatrix} = \frac13(2^n + (1 +\omega)^n + (1 + \omega^2)^n) {/eq}.

## Properties of Cube Root of Unity

The cube root of unity {eq}\omega {/eq} is the root of the equation {eq}z^3 = 1 {/eq}. It has the following basic properties.

• {eq}(\omega)^3 = 1 {/eq}, since {eq}\omega {/eq} is the root of the equation {eq}z^3 = 1 {/eq}.
• Since {eq}(\omega)^3 -1 = 0 {/eq}, we have {eq}(\omega - 1)( (\omega)^2 + \omega +1) = 0 {/eq}. If {eq}\omega \neq 1 {/eq} then {eq}(\omega)^2 + \omega +1 = 0 {/eq}.

## Answer and Explanation:

Firstly, {eq}1 + {\omega}^k + {\omega}^{2k} = 3 {/eq} if {eq}k {/eq} is a multiple of {eq}3 {/eq} otherwise {eq}1 + {\omega}^k + {\omega}^{2k} = 0 {/eq}.

Let {eq}n = 3l + v {/eq} where {eq}v = \{0,1,2\} {/eq}.

We know that

• {eq}(1+1)^n = \sum_{r = 0}^{r = n} {n\choose r} {1}^r \,\,\Rightarrow\, 2^n = \sum_{r = 0}^{r = n} {n\choose r} {/eq}
• {eq}(1+\omega)^n = \sum_{r = 0}^{r = n} {n\choose r} {\omega}^r {/eq}
• {eq}(1+{\omega}^2)^n = \sum_{r = 0}^{r = n} {n\choose r} {\omega}^{2r} {/eq}.

When we add these three expansions, only those terms for which {eq}r = 3k,\,\,\,k = 0,1,2,3..... {/eq} remain and the other terms reduce to zero. For each {eq}r= 3k,\,\,\,k = 0,1,2,3..... {/eq}, we know that {eq}1 + {\omega}^r + {\omega}^{2r} = 3 {/eq}. This is because of the property stated at the beginning of this answer section.

Hence, on adding, we have {eq}\displaystyle 2^n+(1+{\omega})^n +(1+{\omega}^2)^n = \sum_{r = 0}^{r =n } {n\choose r}( 1 + {\omega}^r + {\omega}^{2r}) = \sum_{t = 0}^{t = l} {n\choose r}(3) {/eq} where {eq}r = 3t {/eq} for each {eq}t {/eq}.

Hence, we have {eq}\displaystyle 3\sum_{t = 0}^{t = l} {n\choose r} = 2^n+(1+{\omega})^n +(1+{\omega}^2)^n {/eq}

Expanding the summation shows that {eq}\begin{pmatrix} n\\ 0 \end{pmatrix} + \begin{pmatrix} n\\ 3 \end{pmatrix} + \begin{pmatrix} n\\ 6 \end{pmatrix} + \cdots + \begin{pmatrix} n\\ 3l \end{pmatrix} = \frac13(2^n + (1 +\omega)^n + (1 + \omega^2)^n) {/eq}