COMEDK · Maths · 5. Sequences and Series
If \(U_{n+1}=3 U_n-2 U_{n-1}\) and \(U_0=2, U_1=3\), then \(U_n\) is equal to
- A \(1-2^n\)
- B \(2^n+1\)
- C \(2^n-1\)
- D \(2^n+2\)
Answer & Solution
Correct Answer
(B) \(2^n+1\)
Step-by-step Solution
Detailed explanation
The given recurrence relation is \(U_{n+1} - 3U_n + 2U_{n-1} = 0\). The characteristic equation is \(r^2 - 3r + 2 = 0\). Factoring the quadratic equation, we get \((r-1)(r-2) = 0\), which gives roots \(r_1 = 1\) and \(r_2 = 2\). The general solution is of the form…
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