1is teh ans 2(n-1)/3
dive teh series int grops of 3 -3 terms i.e
(1/(2-w) + 1/(2-w^2) + 1/(2-w^3))=12/7
(1/(2-w^4) + 1/(2-w^5) + 1/(2-w^6))=12/7
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so we get (n-1)/3 groups like tat
value of each group is 2
so ans(4/7)(n-1)
9 Answers
19sensing something to deal with taking logarithm and then a differentiation !
119getting something weird....can be a wrong one ;)
{1+ 2.2 + 3.22 +.......(n-1)2n-2 } /2n
1debo ur hint was perfect , but ....see
log(xn-1)=log((x-1)(x-w)(x-w2)........(x-wn-1))
=log(x-w) +log (x-w2) +..........log(x-wn-1)
now diiff both sides
nxn-1/xn -1 = f(x)
f(2) = is our required summation
1oh wat the helll i did.........i thought w to be cube root of unity.....
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