Consider the recurrence relation \(a_1 = 8, a_n = 6n^2 +2n+a_{n−1}\). Let…
2016
Consider the recurrence relation \(a_1 = 8, a_n = 6n^2 +2n+a_{n−1}\). Let \(a_{99} = K ×10^4\) . The value of \(K\) is ________ .
Attempted by 19 students.
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Correct answer: 197.9 to 198.1
Key insight: The recurrence adds the expression 6n^2 + 2n at each step, and since a1 = 8 equals the value of 6·1^2 + 2·1, we have a_n = sum from k = 1 to n of (6k^2 + 2k).
Compute the required sums for n = 99:
Sum of k from 1 to 99 = 99 × 100 / 2 = 4,950.
Sum of k^2 from 1 to 99 = 99 × 100 × 199 / 6 = 328,350.
Therefore sum of 6k^2 = 6 × 328,350 = 1,970,100, and sum of 2k = 2 × 4,950 = 9,900.
Hence a_99 = 1,970,100 + 9,900 = 1,980,000, which is 198 × 10^4. So K = 198.
Note: The previously provided numeric range (197.9) is incorrect; the exact value is 198.