Question

If ${\alpha }_1,{\alpha }_2,\cdots {\alpha }_{100}$ are all the 100th roots of unity, then $\sum \sum {\left({\alpha }_i{\alpha }_j\right)}^5$ is $1\le i<j\le 100$

• A. $20$
• B. ${\left(20\right)}^{1/20}$
• C. $0$
• D. none

Answer 1

The correct option is C $0$

$2\sum ab={(\sum a)}^2-\sum a^2$
$\therefore 2\sum \sum {\left({\alpha }_i{\alpha }_j\right)}^5={({\alpha }_1^5+{\alpha }_2^5+\cdots )}^2-({\alpha }_1^{10}+{\alpha }_2^{10}+\cdots )$
$=0-0$ ($\because \sum {\alpha }_i^r=100$ if $r=100k$
and $\sum {\alpha }_i^r=0$ if $r\ne 100k$)
Here both 5 and 10 are not multiples of 100.