Let α=−1+i32.if a=(1+α)∑k=0100 α2k&b=∑k=0100 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
Itisgiventhat α=−1+i32, then α2+α+1=0 and α3=1 So, a=(1+α)∑k=0100 α2k=−α2∑k=0100 α2k=−∑k=0100 α2(k+1) =−α2+α4+α6+α8+…+α202 =−α2α2101−1α2−1 (sum of GP) =−α2α202−1α2−1=−α2(α−1)α2−1, ∵α3=1 =−α2α+1=−α3α2+α=1 and, b=∑k=0100 α3k=∑k=0100 α3k=∑k=0100 1=101Now, equation of quadratic equation having roots to'a' and 'b' isx2−(a+b)x+ab=0⇒x2−102x+101=0Hence, option (c) is correct.