Richtige Herleitung:
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- 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 vereinfacht korrekt:
- Homogene Lösung: $a_n^{(h)} = A \cdot 2^n$
- Partikuläre Lösung: Konstant $C$, einsetzen: $C = 2C - 3 \Rightarrow C = 3$
- Allgemeine Lösung: $a_n = A \cdot 2^n + 3$
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