Let α=(-1+i√3/2). If a=(1+α) displaystyle ∑k=0100α2k and displaystyle ∑k=0100α3k, then a and b are the roots of the quadratic equation :

Question

Let α=(-1+i√3/2). If a=(1+α) displaystyle ∑k=0100α2k and displaystyle ∑k=0100α3k, then a and b are the roots of the quadratic equation : (A) x2 + 101x+100=0 (B) x2 + 102x + 101=0 (C) x2 - 102x + 101=0 (D) x2 - 101x + 100=0

Tardigrade Admin · Accepted Answer

α = ω a = (1 + ω )(1 + ω 2 + ω 4 + ..... + ω 200) a = (1+ω ) ((1-(ω 2)101)/1-ω 2) = 1 b = 1+ω 3+ω 6+ ldots ldots+ω 300 = 101 x2 - 102x + 101 = 0

Q. Let $α = \frac{- 1 + i 3}{2} .$ If
$a = (1 + α)$ $k = 0 \sum 100 α^{2 k}$ and $k = 0 \sum 100 α^{3 k}$ , then a and b are the roots of the quadratic equation :

$x^{2} + 101 x + 100 = 0$

$x^{2} + 102 x + 101 = 0$

$x^{2} - 102 x + 101 = 0$

$x^{2} - 101 x + 100 = 0$

$α = ω$
$a = (1 + ω) (1 + ω^{2} + ω^{4} + ..... + ω^{200})$
$a = (1 + ω) \frac{( 1 - ( ω ^{2} ) ^{101} )}{1 - ω ^{2}} = 1$
$b = 1 + ω^{3} + ω^{6} + \dots\dots + ω^{300} = 101$
$x^{2} - 102 x + 101 = 0$

Q. Let α=2−1+i3​​. If a=(1+α) k=0∑100​α2k and k=0∑100​α3k, then a and b are the roots of the quadratic equation :

x2+101x+100=0

x2+102x+101=0

x2−102x+101=0

x2−101x+100=0