Sequences & Series
Recursive sequences
Grade 11
Question:
<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: Define a substitution f(x) = x³ + 2x to recognize that Tₙ₊₁ = f(Tₙ), then analyze properties of this function (monotonicity, fixed points, and iteration behavior) to determine sequence behavior without computing explicit terms.
<p><strong>Step 1:</strong> Define f(x) = x³ + 2x. The recurrence becomes Tₙ₊₁ = f(Tₙ) with T₁ = 1.</p><p><strong>Step 2:</strong> Analyze f(x): f'(x) = 3x² + 2 > 0 for all x ∈ ℝ, so f is strictly increasing.</p><p><strong>Step 3:</strong> For x > 0: f(x) = x³ + 2x > x (since x³ + x > 0). Thus f(x) > x for all x > 0.</p><p><strong>Step 4:</strong> Since T₁ = 1 > 0 and f(Tₙ) > Tₙ for each n, the sequence {Tₙ} is strictly increasing: 1 < T₂ < T₃ < T₄ < ...</p><p><strong>Step 5:</strong> Check boundedness: Since Tₙ₊₁ = Tₙ³ + 2Tₙ ≥ Tₙ³ for Tₙ ≥ 1, and Tₙ grows faster than any linear rate, the sequence is unbounded.</p><p><strong>Step 6:</strong> Therefore: Tₙ is strictly increasing (Option B: ✓), Tₙ → ∞ as n → ∞ (Option D: ✓), and Tₙ is not bounded above (negates other options).</p><p>∴ Answer: B,D</p>
Correct Answer: B,D