If 1, α1, α2, ldots ldots, α2008 are (2009 roots of unity, then the value of Sigma r =12008 r (α r +α2009- r ) equals:

Question

If 1, α1, α2, ldots ldots, α2008 are (2009 roots of unity, then the value of Sigma r =12008 r (α r +α2009- r ) equals: (A) 2009 (B) 2008 (C) 0 (D) -2009

Tardigrade Admin · Accepted Answer

displaystyle ∑ r = 12008r[α r + α 2009 - r]=1[α 1 + α 2008]+2[α 2 + α 2007]+ ldots ldots ldots +2008[α 2008 + α 1] =2009[α 1 + α 2 + α 3 + ldots ldots α 2008]=-2009

Q. If $1, α_{1}, α_{2}, \dots\dots, α_{2008}$ are (2009 roots of unity, then the value of $Σ_{r = 1}^{2008} r (α_{r} + α_{2009 - r})$ equals:

$2009$

$2008$

$0$

$- 2009$

$r = 1 \sum 2008 r [α_{r} + α_{2009 - r}] = 1 [α_{1} + α_{2008}] + 2 [α_{2} + α_{2007}] + \dots\dots\dots + 2008 [α_{2008} + α_{1}]$
$= 2009 [α_{1} + α_{2} + α_{3} + \dots\dots α_{2008}] = - 2009$

Q. If 1,α1​,α2​,……,α2008​ are (2009 roots of unity, then the value of Σr=12008​r(αr​+α2009−r​) equals:

2009

2008

0

−2009

r=1∑2008​r[αr​+α2009−r​]=1[α1​+α2008​]+2[α2​+α2007​]+………+2008[α2008​+α1​] =2009[α1​+α2​+α3​+……α2008​]=−2009

Q. If $1, α_{1}, α_{2}, \dots\dots, α_{2008}$ are (2009 roots of unity, then the value of $Σ_{r = 1}^{2008} r (α_{r} + α_{2009 - r})$ equals:

$2009$

$2008$

$0$

$- 2009$

$r = 1 \sum 2008 r [α_{r} + α_{2009 - r}] = 1 [α_{1} + α_{2008}] + 2 [α_{2} + α_{2007}] + \dots\dots\dots + 2008 [α_{2008} + α_{1}]$
$= 2009 [α_{1} + α_{2} + α_{3} + \dots\dots α_{2008}] = - 2009$