If (S_n=n(6n-1)), find the sum from the (31)st term to the (50)th term.

Correct answer: (9580)

Here \(S_n=n(6n-1)\) represents the sum of the first \(n\) terms. A range beginning at the 31st term is found by taking the total through the 50th term and removing the total through the 30th term. This works because \(S_{30}\) contains precisely the terms that come before the required range.

Evaluate \(S_{50}=50(6\cdot50-1)=50(299)=14950\). Next, \(S_{30}=30(6\cdot30-1)=30(179)=5370\). Thus the sum from the 31st through the 50th term is \(S_{50}-S_{30}=14950-5370=9580\). Therefore option D is correct. The number of included terms is 20, from 31 to 50 inclusive, which is consistent with this subtraction method.