If represents a cube root of unity and ∑ k = 1 n k + 1 ω k + 1 ω 2 = 340 , then n =

Step-by-step solution

Given, ∑ k = 1 n k + 1 ω k + 1 ω 2 = 340 ∑ k = 1 n 1 + 1 ω 1 + 1 ω 2 + 2 + 1 ω 2 + 1 ω 2 + . . . . . . . . + n + 1 ω n + 1 ω 2 = 340 Now, ∑ k = 1 n k + 1 ω k + 1 ω 2 = 340 ∑ k = 1 n k 2 + k ω + k ω 2 + 1 ω 3 = 340 ∑ k = 1 n k 2 + k 1 ω + 1 ω 2 + 1 ω 3 = 340 ∑ k = 1 n k 2 + k ω + 1 ω 2 + 1 ω 3 = 340 We know that ω 3 = 1 ,   1 + ω + ω 2 = 1 ∑ k = 1 n k 2 + k - &#