For each positive integer n , let S_n denote the sum of the infinite series 4 + 8 (n+1)^2 + 12 (n+1)^4 + 16 (n+1)^6 + . The value of 231 _ n=1 ²⁰ ( S_n - 2 ) is equal to

Step-by-step solution

The given series is an infinite Arithmetic-Geometric Progression (AGP) with first term a = 4 , common difference of the arithmetic part d = 4 , and common ratio of the geometric part r = 1 (n+1)^2 . Let S_n = 4 + 8r + 12r^2 + 16r^3 + Multiply by r : rS_n = 4r + 8r^2 + 12r^3 + Subtracting the two equations: (1-r)S_n = 4 + 4r + 4r^2 + 4r^3 + (1-r)S_n = 4 1-r S_n = 4 (1-r)^2 Substitute r = 1 (n+1)^2 : 1-r = 1 - 1 (n+1)^2 = n^2+2n (n+1)^2 = n(n+2) (n+1)^2 Thus, S_n = 4(n+1)^4 n^2(n+2)^2 . Taking the square root: S_n =