Let a n n = 0 ∞ be a sequence such that a 0 = a 1 = 0 and a n + 2 = 2 a n + 1 - a n + 1 for all n ≥ 0 . Then, ∑ n = 2 ∞ a n 7 n is equal to

Step-by-step solution

Given a n + 2 = 2 a n + 1 - a n + 1 Dividing by 7 n + 2 , we get a n + 2 7 n + 2 = 2 7 · a n + 1 7 n + 1 - 1 49 · a n 7 n + 1 7 n + 2 So, ∑ n = 2 ∞ a n + 2 7 n + 2 = 2 7 ∑ n = 2 ∞ a n + 1 7 n + 1 - 1 49 ∑ n = 2 ∞ a n 7 n + ∑ n = 2 ∞ 1 7 n + 2 Let ∑ n = 2 ∞ a n 7 n = P ⇒ P - a 3 7 3 - a 2 7 2 = 2 7 P - a 2 7 2 - 1 49 P + 1 7 4 1 1 - 1 7 ∵ a 2 = 1   &   a 3 = 3 ⇒ P 1 - 2 7 + 1 49 = 3 7 3 + 1 7 2 - 2 7 3 + 1 6 ·