Ta có:
6 k 3 k + 1 - 2 k + 1 3 k - 2 k = 6 3 k - 2 k 3 k + 1 - 2 k + 1 - 3 k + 1 - 2 k + 1 3 k - 2 k ∑ k = 1 n 6 k 3 k + 1 - 2 k + 1 3 k - 2 k = 6 3 n - 2 n 3 n + 1 - 2 n + 1
Do đó:
lim x → ∞ ∑ k = 1 n 6 k 3 k + 1 - 2 k + 1 3 k - 2 k = 6 lim n → ∞ 3 n - 2 n 3 n + 1 - 2 n + 1 = 6 lim n → ∞ 1 - 2 3 n 1 - 2 . 2 3 2 = 2
Đáp án D