1Someone please help !!!
If a1, a2, a3, a4,..........., a100 are the 100 roots of unity then find the value of
\sum_{1\leq i \leq j \leq 100}^{}{} (aiaj)5 .
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4 Answers
1Let a1 be θ.
Thus a2=θ2,a3=θ3,......a100=θ100
Now θ100=1
Notice that, if anis a root of unity, then (an)5 is also a root of unity.
Thus from theory of equations we have
\sum_{1\leq\i\leq j\leq 100 }^{}{}(aiaj)5 = \sum_{1\leq i<j\leq 100}^{}{}(aiaj)5 +\sum_{i=1}^{100}{}(ai)10 = 0..
I hope thats the answer.
1The eqn. whose roots are a1, a2, a3, a4....., a100 is x100-1=0
1i≤j≤100, gives the sum of roots taken two at a time + a12+a22+a32+a42........+a1002
From the eqn, sum of roots taken two at a tym is 0
Taking, ar=ei∩r/100
ap2=a2p
Which gives, a12+a22+a32+a42........+a1002 = 2 (a2+a4+a6........+a100)
These are roots of x50-1=0 and add upto 0
Hence the given expression,
\sum_{1\leq i\leq j\leq 100}^{}{}(aiaj)5 = 05= 0
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