Let i = √(- 1). If ( - 1 + i√3 )²¹/(1 - i)²⁴ + ( 1 + i√3 )²¹/(1 + i)²⁴ = k, and n = [|k|] be the greatest integral part of | k |. Then Σⱼ₌₀ⁿ⁺⁵ (j + 5)² - Σⱼ₌₀ⁿ⁺⁵ (j + 5) is equal to _________.

Solution

Maths sir · 4 steps

Write each base as r e^(iθ); the two big fractions become +512 and -512, so k = 0.

1. Polar forms. -1 + i√3 = 2e^(i2π/3), 1 + i√3 = 2e^(iπ/3), 1 - i = √2 e^(-iπ/4) and 1 + i = √2 e^(iπ/4). De Moivre's theorem says (re^(iθ))ᵐ = rᵐ e^(imθ).

   2²¹e^(i14π)/2¹²e^(-i6π) + 2²¹e^(i7π)/2¹²e^(i6π) = 2⁹(1) + 2⁹(-1) = 0

2. Find n. e^(i14π) = 1, e^(i7π) = -1 and e^(± i6π) = 1, so the terms are 512 and -512. Hence k = 0, |k| = 0 and n = 0.
3. Evaluate the sums. With n = 0, j runs from 0 to 5, so j + 5 runs from 5 to 10. The squares add to 25 + 36 + 49 + 64 + 81 + 100 = 355, and the numbers add to 5 + 6 + 7 + 8 + 9 + 10 = 45.
4. So. The required value is 355 - 45 = 310.

The common slip: 290. Starting the sums at j = 1 drops the j = 0 term 25 - 5 = 20 and gives 290.

The idea to keep. For big powers of 1 ± i or ± 1 ± i√3, switch to re^(iθ) at once. A multiple of π in the angle turns the answer into a plain ± real number.

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