For each integer k = 1, 2, , 10 , let S_k be the sum of an infinite geometric progression whose first term is k^2 and whose common ratio is k-1 k+1 . Then the value of _ k=1 ¹⁰ S_k is equal to

Step-by-step solution

The sum of an infinite geometric progression with first term a and common ratio r is given by S = a 1-r . For the k -th geometric progression, a = k^2 and r = k-1 k+1 . Thus, S_k = k^2 1 - k-1 k+1 S_k = k^2 k+1 - (k-1) k+1 S_k = k^2(k+1) 2 = 1 2 (k^3 + k^2) We need to find the sum _ k=1 ¹⁰ S_k : _ k=1 ¹⁰ S_k = 1 2 _ k=1 ¹⁰ (k^3 + k^2) _ k=1 ¹⁰ S_k = 1 2 [ _ k=1 ¹⁰ k^3 + _ k=1 ¹⁰ k^2 ] Using the standard summation formulas _ k=1 ^n k^3 = ( n(n+1) 2 )^2 and _ k=1 ^n k^2 = n(n+1)(2n+1) 6 for n=10 : _ k=1 ¹⁰ k^3 = ( 10