Question

If ${\alpha }_1,{\alpha }_2,{\alpha }_3,......,{\alpha }_{100}$ are all the $100$th roots of unity. Then the numerical value of $\underset{1\le i<j\le 100}{\sum \sum }({\alpha }_i{\alpha }_j{)}^5$ is

Answer 1

We know that
$2\underset{1\le i<j\le 100}{\sum \sum }({\alpha }_i{\alpha }_j{)}^5=\sum _{i=1}^{100}\sum _{j=1}^{100}({\alpha }_i{\alpha }_j{)}^5-\sum _{i=1}^{100}(a_i{)}^{10}$
$\therefore 2\underset{1\le i<j\le 100}{\sum \sum }({\alpha }_i{\alpha }_j{)}^5={({\alpha }_1^5+{\alpha }_2^5+...+({\alpha }_{100}{)}^5)}^2-{({\alpha }_1^{10}+{\alpha }_2^{10}+...+({\alpha }_{100})}^{10})$
$=0-0=0$
[$\because ({\alpha }_1^r+{\alpha }_2^r+{\alpha }_3^r+......+{\alpha }_{100}^r)=0$
if $r$ is not a multiple of $100,\because {\alpha }^{100}=1$]