a nice question .....a doubt

sensing something to deal with taking logarithm and then a differentiation !

is 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
.
.
.
.
.

so we get (n-1)/3 groups like tat

value of each group is 2
so ans(4/7)(n-1)

solve it ;)

is this teh ans (4/7)(n-1)

getting something weird....can be a wrong one ;)

{1+ 2.2 + 3.2² +.......(n-1)2^n-2 } /2ⁿ

debo ur hint was perfect , but ....see
log(xⁿ-1)=log((x-1)(x-w)(x-w²)........(x-w^n-1))
=log(x-w) +log (x-w²) +..........log(x-w^n-1)
now diiff both sides
nx^n-1/xⁿ -1 = f(x)
f(2) = is our required summation

yes, i did just that ! [1] [1]