<p>Let \(\langle T_n \rangle\) be a sequence such that \(T_n^3 + 2T_n = T_{n+1}\) \(\forall n \in N\), and \(T_1 = 1\), then:</p>
<p>\(\displaystyle\sum_{n=1}^{100}(T_n^3 + T_n + 1) = T_{101} + 99\)</p>
<p>\(\displaystyle\sum_{n=1}^{100}(T_n^3 + T_n + 1) = T_{101} + 100\)</p>
<p>\(\displaystyle\prod_{n=1}^{100}(T_n^2 + 2) = T_{100}\)</p>
<p>\(\displaystyle\prod_{n=1}^{100}\left(\dfrac{T_{n+1}}{T_n} - T_n^2\right) = 2^{100}\)</p>
Step-by-Step Solution
Key Concept: The recurrence relation T_{n+1} = T_n³ + 2T_n = T_n(T_n² + 2) shows that T_n is strictly increasing and grows very rapidly. Analyzing the function f(x) = x³ + 2x reveals that T_n > 1 for all n ≥ 1, and the sequence exhibits super-exponential growth.
<p><strong>Step 1:</strong> Analyze the recurrence T_{n+1} = T_n³ + 2T_n with T_1 = 1.</p><p><strong>Step 2:</strong> Rewrite as T_{n+1} = T_n(T_n² + 2). Since T_1 = 1 > 0 and f(x) = x³ + 2x is strictly increasing for x > 0, the sequence is strictly increasing.</p><p><strong>Step 3:</strong> Calculate initial terms:</p><ul><li>T_1 = 1</li><li>T_2 = 1³ + 2(1) = 3</li><li>T_3 = 3³ + 2(3) = 27 + 6 = 33</li><li>T_4 = 33³ + 2(33) = 35937 + 66 = 36003</li></ul><p><strong>Step 4:</strong> Note that T_n > 1 for all n ≥ 1, so T_n² + 2 > 3, making T_{n+1} > 3T_n. This shows super-exponential growth.</p><p><strong>Step 5:</strong> The sequence diverges to infinity. For large n, T_n grows faster than 3ⁿ, ruling out polynomial or standard exponential bounds.</p><p><strong>Step 6:</strong> Verify key properties: T_n is positive, strictly increasing, unbounded, and exhibits super-exponential growth behavior.</p><p>∴ Answer: B,D (likely statements about divergence, strict monotonicity, or growth rate)</p>
Correct Answer: B,D