Consider the sequence a 1 , a 2 , a 3 , … … such that a 1 = 1 , a 2 = 2 and a n + 2 = 2 a n + 1 + a n for n = 1 , 2 , 3 , … If a 1 + 1 a 2 a 3 · a 2 + 1 a 3 a 4 · a 3 + 1 a 4 a 5 … a 30 + 1 a 31 a 32 = 2 α C 31 61 then α is equal to

Step-by-step solution

Given a n + 2 = 2 a n + 1 + a n ⇒ a n + 2 a n + 1 - a n a n + 1 = 2 Now, T r = a r + 1 a r is an A.P. with common difference 2 Now, T 1 = a 1 a 2 = 2 ,   T 2 = a 2 a 3 = 4 ,   . . . . ,   T r = 2 r So a 1 + 1 a 2 a 3 · a 2 + 1 a 3 a 4 ⋯ a 30 + 1 a 31 a 32 = ∏ i = 1 30 a i + 1 a i + 1 a i + 2 = ∏ i = 1 30 a i a i + 1 + 1 a i + 1 a i + 2 = ∏ i = 1 30 T r + 1 T r + 1 = ∏ i = 1 30 2 r + 1 2 r + 2 = 3 · 5 · 7 ⋯ 61 4 · 6 · 8 ⋯ 62 = 1