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

Solution

Maths sir · 4 steps

Recognise α as ω, then use ω³ = 1 to collapse both sums into plain numbers.

1. Spot omega. α = (-1 + i√3)/2 is the cube root of unity ω. So ω³ = 1, 1 + ω + ω² = 0 and 1 + ω = -ω².
2. Sum for a. The powers ω²ᵏ repeat as 1, ω², ω every 3 terms, and each block adds to 0. The 101 terms make 33 blocks (k = 0 to 98) plus two leftover terms, ω¹⁹⁸ = 1 and ω²⁰⁰ = ω².

   a = (1+ω)(1 + ω²) = (-ω²)(-ω) = ω³ = 1

3. Sum for b. α³ᵏ = (ω³)ᵏ = 1 for every k. There are 101 terms (k = 0 to 100), so b = 101.
4. So. The roots are 1 and 101: sum 102, product 101. The equation is x² - 102x + 101 = 0, option (c).

The idea to keep. Whenever you see (-1 ± i√3)/2, write it as ω and reduce every power modulo 3. Long sums of powers of ω then shrink to the few leftover terms.

Answer from the NTA final answer key. Solution by PrepWiser.