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Solving for \(x\), we get: - Ready Digital AB

📅 March 8, 2026 👤 scraface

Mar 08, 2026

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B: $a_n = 2^{n+1} - 3$
C: $a_n = 5 \cdot 2^n - 3$
D: $a_n = 2^{n} \cdot 5 - 3 \cdot \frac{2^n - 1}{1}$
Antwort: D $a_n = 2^{n} \cdot 5 - 3 \cdot \frac{2^n - 1}{1}$ — jedoch...
Richtige Herleitung:
Homogene Lösung: $a_n^{(h)} = A \cdot 2^n$
Partikuläre Lösung: Konstant $C$, einsetzen: $C = 2C - 3 \Rightarrow C...
Allgemeine Lösung: $a_n = A \cdot 2^n + 3$
Nutze $a_0 = 5$: $5 = A + 3 \Rightarrow A = 2$
Also: $a_n = 2 \cdot 2^n + 3 = 2^{n+1} + 3$? Warte — Prüfe:

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