For any positive n , let f n = 4 n + 4 n 2 - 1 2 n + 1 + 2 n - 1 . Then, ∑ k = 1 40 f k 100 is equal to

Step-by-step solution

Let, x = 2 n + 1 and y = 2 n - 1 ⇒ x 2 + y 2 = 4 x x 2 - y 2 = 2 Also, x y = 4 n 2 - 1 So, f n = x 2 + y 2 + x y x + y = x 3 - y 3 x 2 - y 2 = x 3 - y 3 2 = 1 2 2 n + 1 3 2 2 n - 1 3 2 Substituting n = 1 to 40, we get f 1 = 1 2 3 3 2 - 1 3 2 f 2 = 1 2 5 3 2 - 3 3 2 f 3 = 1 2 7 3 2 - 5 3 2 f 40 = 1 2 81 3 2 - 79 3 2 ⇒ ∑ n = 1 40 = 1 2 81 3 2 - 1 3 2 = 729 - 1 2 = 364