Question

Suppose n is a natural number such that | i + 2i² + 3i^3+
... + ni^n | = 18√2, where i is the square root of -1. Then
n is​

Answer 1

i + 2i² + 3i³ + ... + niⁿ = 18√2

1 + 2i + 3i² + ... + niⁿ⁻¹ = (18√2)/i

f(x) = 1 + 2x + 3x² + ... + nxⁿ⁻¹

∫ f(x) dx = Σₖ₌₁ⁿ (xᵏ)

∫ f(x) dx = x + x² + x³ + ... + xⁿ

∫ f(x) dx = x·(xⁿ - 1)/(x-1)

∫ f(x) dx = (xⁿ⁺¹ - x)/(x-1)

f(x) = [(n+1)xⁿ - 1][(x-1) - (xⁿ⁺¹ - x)]/(x-1)²

f(x) = [(n+1)xⁿ⁺¹ - (n+1)xⁿ - x + 1 - xⁿ⁺¹ + x]/(x-1)²

f(x) = [nxⁿ⁺¹ - (n+1)xⁿ + 1]/(x-1)²

⇒ f(i) = 18√2

⇒ [niⁿ⁺¹ - (n+1)iⁿ +1]/(i-1)² = 18√2

⇒ niⁿ⁺¹ - (n+1)iⁿ + 1 = -36i√2

⇒ n ∈ Ф