double sigma in complex number if $\beta_{1},\beta_{2},\cdots ,\beta_{100}$ are $100$ th roots of unity. The numerical value of $\displaystyle \mathop{\sum \sum}_{1\leq i\leq j\leq 100}\bigg(\beta_{i}\beta_{j}\bigg)^5$ is 
what i try
assuming $x=(1)^{\frac{1}{100}},$ then $x^{100}-1=0$ has roots
$x=\beta_{1},\beta_{2},\cdots,\beta_{100}$
$\displaystyle x^{100}-1=(x-\beta_{1})(x-\beta_{2})\cdots \cdots (x-\beta_{100})$
how do i solve it help me please 
 A: 
We   obtain
  \begin{align*}
\color{blue}{\sum_{1\leq  i\leq     j\leq   100}}\color{blue}{\left(\beta_i\beta_j\right)^5}
&=\sum_{j=1}^{100}\beta_j^5\sum_{i=1}^j\beta_i^5\tag{1}\\
&=\sum_{j=1}^{100}\beta_1^{5j}\sum_{i=1}^j\left(\beta_1^5\right)^i\tag{2}\\
&=\sum_{j=1}^{100}\beta_1^{5j}\frac{\beta_1^5-\left(\beta_1^5\right)^{j+1}}{1-\beta_1}\tag{3}\\
&=\frac{\beta_1^5}{1-\beta_1}\left(\sum_{j=1}^{100}\beta_1^{5j}-\sum_{j=1}^{100}\beta_1^{10j}\right)\tag{4}\\
&=\frac{\beta_1^5}{1-\beta_1}\left(\frac{\beta_1^5-\beta_1^{505}}{1-\beta_1^5}-\frac{\beta_1^{10}-\beta_1^{1010}}{1-\beta_1^{10}}\right)\tag{5}\\
&\,\,\color{blue}{=0}
\end{align*}

Comment:


*

*In (1) we rewrite the sum as double sum and factor out $\beta_j$.

*In (2) we use $\beta_j=\exp\left(\frac{2\pi i j}{100}\right)=\exp\left(\frac{2\pi i}{100}\right)^j=\beta_1^j$.

*In (3) we apply the finite geometric series formula.

*In (4) we factor out common terms and multiply out the sums.

*In (5) we again apply the finite geometric series formula. We  observe, since $\beta_1^{100}=1$ that $\beta_1^{505}=\beta_1^5$ and $\beta_1^{10}=\beta_1^{1010}$.
Note:
Since we have $\beta_1^{100}-1=(\beta_1-1)\left(1+\beta_1+\cdots+\beta_1^{99}\right)=0$ we  can deduce   the result      zero   already    from (4).
