Explanation

This is a mean-of-transformed-data problem. The new numbers are i(i+1) = i² + i for i = 1, 2, …, n, so their sum is Σi² + Σi. Using the standard summation formulas Σi² = n(n+1)(2n+1)/6 and Σi = n(n+1)/2, the total is n(n+1)(2n+1)/6 + n(n+1)/2 = n(n+1)(n+2)/3. The mean is this sum divided by the count n, which gives (n+1)(n+2)/3. Hence option (d). Distractors like (n+1)(n+2)/3n wrongly divide by n twice, and n(n+1)/2 is just the sum of the original unweighted numbers.