Let a n n = 0 ∞ be a sequence such that a 0 = a 1 = 0 and a n + 2 = 3 a n + 1 - 2 a n + 1 , ∀ n ≥ 0 . Then a 25 a 23 - 2 a 25 a 22 - 2 a 23 a 24 + 4 a 22 a 24 is equal to

Step-by-step solution

Given, a 0 = 0 , a 1 = 0 And a n + 2 = 3 a a + 1 - 2 a n + 1 : n ≥ 0 ⇒ a n + 2 - a n + 1 = 2 a n + 1 - a n + 1 Now for n = 0    a 2 - a 1 = 2 a 1 - a 0 + 1   . . . . 1 For n = 1    a 3 - a 2 = 2 a 2 - a 1 + 1   . . . . . . . . 2 For n = 2    a 4 - a 3 = 2 a 3 - a 2 + 1   . . . . . . . . 3 . . . For n = n    a n + 2 - a n + 1 = 2 a n + 1 - a n + 1   . . . . . . n Now adding all above equation upto n we get, a n + 2 - a 1 - 2 a n + 1 - a 0 - n +