Question Number 148501 by mathmax by abdo last updated on 28/Jul/21

let U_n ={z∈C /z^n =1} simplify Σ_(p=0) ^(2n−1) w^p with w∈U_n and Σ_(p=0) ^(2n−1) (2w +1)^p

Answered by Olaf_Thorendsen last updated on 28/Jul/21

• w = 1 S_n (1) = Σ_(p=0) ^(2n−1) 1^p = 2n (trivial) • w^n = 1 and w ≠ 1 S_n (w) = Σ_(p=0) ^(2n−1) w^p =((1−w^(2n) )/(1−w)) = ((1−1)/(1−w)) = 0 T_n (w) = Σ_(p=0) ^(2n−1) (2w+1)^p T_n (w) =((1−(2w+1)^(2n) )/(1−(2w+1))) = (((2w+1)^(2n) −1)/(2w))

Answered by mathmax by abdo last updated on 28/Jul/21

if w=1 A_n =Σ_(p=0) ^(2n−1) w^p =2n if w≠1 Σ_(p=0) ^(2n−1) w^p =((1−w^(2n) )/(1−w))=((1−(w^n )^2 )/(1−w))=((1−1)/(1−w))=0 alsoB_n = Σ_(p=0) ^(2n−1) (2w+1)^p the roots of z^n =1 are z_k =e^(i((2kπ)/n)) and 0≤k≤n−1 ⇒w≠−(1/2) ⇒B_n =((1−(2w+1)^(2n) )/(1−(2w+1))) =(1/(2w)){(2w+1)^(2n) −1} =(1/(2w)){Σ_(k=0) ^(2n) C_(2n) ^k (2w)^k −1} =(1/(2w))Σ_(k=0) ^(2n) C_(2n) ^k 2^k w^k