The sum of n terms of the series 4 + 44 + 444 + ... is: A. (4/81)(10^(n+1) - 9n - 10) B. (4/81)(10^n - 9n - 1) C. (4/81)(10^(n+1) - 9n - 10) D. (4/81)(10^n - 9n - 16)

GATE 2011 · General Aptitude · Arithmetic Series · medium

Answer: S_n = (4/81)(10^(n+1) - 9n - 10)

1. Express k-th term: T_1 = (4/9)(10-1) = 4, T_2 = (4/9)(100-1) = 44, T_3 = (4/9)(999) = 444. General term verified.
2. Sum all n terms: S_n = (4/9)[(10 + 10^2 + ... + 10^n) - n]
3. Evaluate the geometric sum: Substitute: S_n = (4/9)[(10^(n+1) - 10)/9 - n]
4. Simplify to final form: S_n = (4/81)(10^(n+1) - 9n - 10). This matches option C.