Each term is (4/9)(10^k − 1) for k = 1,2,...,n (e.g. 44 = (4/9)(99) = (4/9)(10²−1)). Sum = (4/9)Σ(10^k − 1) = (4/9)[Σ10^k − n] = (4/9)[(10^(n+1)−10)/9 − n] = (4/81)[10^(n+1) − 10 − 9n] = (4/81)[10^(n+1) − 9n − 10], matching option (C) exactly (verified numerically for n=5: series sum = 49380, and the formula gives the same value).
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