Home
Class 12
MATHS
If omega(!= 1) is a cube root of unity, ...

If `omega(!= 1)` is a cube root of unity, then `1/sqrt3|1+2omega+3omega^2+..........+3nomega^(3n-1)|(n in N)` cannot exceed

Promotional Banner

Similar Questions

Explore conceptually related problems

If omega ne1 is a cube root of unity then the sum of the series S=1+2omega+3omega^2+….+3nomega^(3n-1) is

If omega(!=1) is a cube root of unity,then the sum of the series S=1+2 omega+3 omega^(2)+....+3n omega^(3n-1) is

If omega is a complex cube root of unity, then (1-omega+(omega)^2)^3 =

If omega is a cube root of unity,then (1+omega)^(3)-(1+omega^(2))^(3)=

If omega ne 1 is cube root of unity, then the sum of the series S = 1+2omega+3omega^2+….......+3n(omega)^(3n-1) is

If omega ne 1 is a cube root of unity, then the sum of the series S = 1+ 2omega + 3omega^(2) +…….+3n omega^(3n-1) is: