${\mathrm{S}}_1\equiv \text{ If }{\mathrm{z}}^2+\mathrm{z}+1=0\text{ and }\mathrm{n}\in \mathrm{N}\text{ then }\sum _{\mathrm{k}=1}^{\mathrm{n}} {\mathrm{z}}^{\mathrm{k}}+{\mathrm{z}}^{-\mathrm{k}}=-\mathrm{n}+3\left[\frac{\mathrm{n}}{3}\right]\text{ where [ ] gif }$

${\mathrm{S}}_2\equiv \text{ If }\mathrm{w}\ne 1\text{ is a cube root of }1\text{ then }{\mathrm{w}}^{\mathrm{k}}+{\mathrm{w}}^{-\mathrm{k}}=\left\{\begin{array}{c}-1\text{ if }\mathrm{k}\text{ is not a multiple of }3\\ 2\text{ if }\mathrm{k}\text{ is a multiple of }3\end{array}\right.\text{ then }$

$\mathrm{w}+{\mathrm{w}}^2=-1,{\mathrm{w}}^3=1$

$$\begin{gathered} \text{ For statement-2, note that }{\omega }^k={\overline{\omega }}^k=1\text{ if }k\text{ is a mul tiple of }3\text{, and if }k\text{ is not a multiple of }3\text{,} \\ \text{then one of }{\omega }^k,{\overline{\omega }}^k\text{ equals }\omega \text{ and other equals }{\omega }^2\text{. } \end{gathered}$$

$\text{ Next, note that roots of }{z}^2-z+1=0\text{ are }-\omega \text{ and }-{\omega }^2=-\overline{\omega }-1/\omega \text{. Now, use }$

${\left(z^k+z^{-k}\right)}^2={\left[(-\omega {)}^k+(-\overline{\omega }{)}^k\right]}^2$

$=\left\{\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }k\text{ is not a multiple of }3\\ 4\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }k\text{ is a multiple of }3\end{array}\right.$