Let for n = 1 , 2 , … … , 50 , S n be the sum of the infinite geometric progression whose first term is n 2 and whose common ratio is 1 n + 1 2 . Then the value of 1 26 + ∑ n = 1 50 S n + 2 n + 1 - n - 1 is equal to

Step-by-step solution

We have, a = n 2 ,   r = 1 n + 1 2 S n = n 2 1 - 1 n + 1 2 = n 2 n + 1 2 n n + 2 = n n n + 2 + 1 n + 2 = n 2 + n n + 2 = n 2 + n + 2 - 2 n + 2 = n 2 + 1 - 2 n + 2               . . . i Now, ∑ n = 1 50 S n + 2 n + 1 - n - 1 From equation i , = ∑ n = 1 50 n 2 + 1 - 2 n + 2 + 2 n + 1 - n - 1 = ∑ n = 1 50 n 2 - n + 2 ∑ n = 1 50 1 n + 1 - 1 n + 2 = ∑ n = 1 50 n 2 - ∑ n = 1 50 n + 2 1 2 - 1 3 + 1 3 - 1 4 + ⋯ + 1 51 - 1 52 = 50 51 101 6 - 50