Consider a function f: N R with f(1) = 3 , satisfying the relation 2 _ k=1 ^x (k+1)f(k) = (x+1)^2 f(x) + 4 for all integers x 2 . Then the value of _ x=2 ²⁰ x+1 x f(x) is equal to

Step-by-step solution

Let S_x = _ k=1 ^x (k+1)f(k) . The given equation is 2S_x = (x+1)^2 f(x) + 4 for x 2 . Replacing x with x-1 , we get 2S_ x-1 = x^2 f(x-1) + 4 for x 3 . Subtracting the two equations yields: 2(x+1)f(x) = (x+1)^2 f(x) - x^2 f(x-1) Rearranging terms, we obtain: x^2 f(x-1) = [(x+1)^2 - 2(x+1)] f(x) = (x^2 - 1) f(x) Thus, f(x) = x^2 x^2-1 f(x-1) for x 3 . To find f(2) , substitute x=2 into the original sum equation: 2[2f(1) + 3f(2)] = 9f(2) + 4 Given f(1) = 3 , we have: 2[6 + 3f(2)] = 9f(2) + 4 12 + 6f(2) = 9f(2) + 4 3f