How many terms of the series 1 + 3 + 5 + 7 + amount to a sum equal to 12345678987654321 ?

Step-by-step solution

The given series is 1 + 3 + 5 + 7 + This is an arithmetic progression with first term a = 1 and common difference d = 2 . The sum of the first n terms of this series is given by: S_n = n 2 [2a + (n-1)d] = n 2 [2(1) + (n-1)2] = n^2 We are given that S_n = 12345678987654321 . Therefore, n^2 = 12345678987654321 . Using the pattern of squares of numbers consisting entirely of ones: 1^2 = 1 11^2 = 121 111^2 = 12321 11 1 _ k times ^2 = 12 k 21 (for k 9 ) Since the maximum digit in the given sum is 9 , the square root con