Sequences & Series
Recurrence Relations
Grade 11
Question:
<p><strong>Statement-1:</strong> If <latex>a_1 = 1, a_2 = 5</latex>, then <latex>a_n = 3^n - 2^n, \forall n \in \mathbb{N}</latex> and <latex>n \geq 1</latex>.</p><p><strong>Statement-2:</strong> <latex>a_{n+2} = 5a_{n+1} - 6a_n, n \geq 1</latex>.</p>
<p>(a) Statement-1 is true, Statement-2 is true; Statement-2 is correct explanation for Statement-1</p>
<p>(b) Statement-1 is true, Statement-2 is true; Statement-2 is not correct explanation for Statement-1</p>
<p>(c) Statement-1 is true, Statement-2 is false</p>
<p>(d) Statement-1 is false, Statement-2 is true</p>
Step-by-Step Solution
Key Concept: Verify both the explicit formula and its recurrence relation; use induction to confirm they are consistent.
<p>Verify: <latex>a_1 = 3^1 - 2^1 = 1</latex> ✓, <latex>a_2 = 3^2 - 2^2 = 5</latex> ✓. The recurrence relation <latex>a_{n+2} = 5a_{n+1} - 6a_n</latex> is the characteristic equation of the solution <latex>a_n = 3^n - 2^n</latex>. By mathematical induction, Statement-1 follows from Statement-2.</p>
Correct Answer: a