Sequences & Series
Miscellaneous
Grade 11
Question:
<p>The 2008th term of the sequence \(1, \underbrace{2,2,2}_{3}, \underbrace{3,3,3,3,3}_{6}, \underbrace{4,4,4,4,4,4,4,4,4,4}_{10}, \ldots\) where \(n\) occurs \(\dfrac{n(n+1)}{2}\) times in the sequence, equals \(k\), then find \(\dfrac{k}{5}\).</p>
Step-by-Step Solution
Key Concept: Find which number's block contains the 2008th term by determining when the cumulative sum of occurrences first exceeds 2008. The number n appears T(n) = n(n+1)/2 times, so we need the smallest n where Σ T(i) ≥ 2008.
<p><strong>Step 1:</strong> Find the cumulative count formula. If number n appears T(n) = n(n+1)/2 times, the total terms up through all occurrences of n is:</p><p>S(n) = Σ(i=1 to n) [i(i+1)/2] = (1/2)Σ(i=1 to n) [i² + i]</p><p>= (1/2)[n(n+1)(2n+1)/6 + n(n+1)/2] = n(n+1)(n+2)/6</p><p><strong>Step 2:</strong> Find n such that S(n-1) < 2008 ≤ S(n).</p><p>For n = 17: S(17) = 17·18·19/6 = 5814/6 = 969</p><p>For n = 18: S(18) = 18·19·20/6 = 6840/6 = 1140</p><p>For n = 19: S(19) = 19·20·21/6 = 7980/6 = 1330</p><p>For n = 20: S(20) = 20·21·22/6 = 9240/6 = 1540</p><p>For n = 21: S(21) = 21·22·23/6 = 10626/6 = 1771</p><p>For n = 22: S(22) = 22·23·24/6 = 12144/6 = 2024</p><p><strong>Step 3:</strong> Since S(21) = 1771 < 2008 < 2024 = S(22), the 2008th term falls in the block of 22s.</p><p>Position within the 22-block: 2008 - 1771 = 237</p><p>Since 22 appears T(22) = 22·23/2 = 253 times and 237 ≤ 253, the 2008th term is indeed 22.</p><p><strong>Step 4:</strong> Therefore k = 22, so k/5 = 22/5 = <strong>4.4</strong> or <strong>44/10</strong></p><p>∴ Answer: <strong>22/5 or 4.4</strong></p>
Correct Answer: 22