Let S_n = _ k=0 ^ n^3 (n+1)^ 2k for n = 1, 2, , 10 . The value of 3 _ n=1 ¹⁰ ( S_n - 4 n+1 ) is equal to

Step-by-step solution

The inner sum is an infinite geometric progression with first term a = n^3 and common ratio r = 1 (n+1)^2 . Sum of the infinite GP is given by: S_n = a 1-r = n^3 1 - 1 (n+1)^2 = n^3(n+1)^2 (n+1)^2 - 1 = n^3(n^2+2n+1) n^2+2n S_n = n^2(n+1)^2 n+2 = n^4+2n^3+n^2 n+2 Performing polynomial long division of n^4+2n^3+n^2 by n+2 : n^4+2n^3+n^2 = n^3(n+2) + n(n+2) - 2(n+2) + 4 Thus, S_n = n^3 + n - 2 + 4 n+2 . Substitute this into the required summation: _ n=1 ¹⁰ ( S_n - 4 n+1 ) = _ n=1 ¹⁰ ( n^3 + n - 2 + 4 n+2 - 4 n+1 ) Th