The sum of the first n terms in the sequence 5, 55, 555, 5555, ... is: A. (5/81)(10^(n+1) - 9n - 10) B. (5/81)(10^(n+1) - 9n - 10) C. (5/81)(10^(n+1) + 9n - 10) D. (5/81)(10^(n+1) - 9n + 10)

GATE 2020 · General Aptitude · Arithmetic Series · medium

Answer: S_n = (5/81)(10^(n+1) - 9n - 10). Answer: B.

1. Express the k-th term: T_1 = 5/9*(10-1) = 5/9*9 = 5. T_2 = 5/9*(100-1) = 55. T_k = (5/9)(10^k - 1). Confirmed.
2. Write the sum and factor out (5/9): S_n = (5/9) * [(10 + 10^2 + ... + 10^n) - n].
3. Evaluate the geometric sum: sum_{k=1}^{n} 10^k = (10^(n+1) - 10)/9.
4. Combine and simplify: S_n = (5/9) * [(10^(n+1) - 10 - 9n)/9] = (5/81) * (10^(n+1) - 9n - 10).