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Bepaal die waarde van die volgende: \( \sum_{n=1}^{\infty} 2^{1-2 n} \)

Ask by Bowen Bolton. in South Africa
Jan 23,2025

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Die waarde van die reeks \( \sum_{n=1}^{\infty} 2^{1-2n} \) is \( \frac{2}{3} \).

Solution

Ons wil die waarde bepaal van die oneindige reeks: \[ \sum_{n=1}^{\infty} 2^{1-2n} \] **Stappe om die som te bereken:** 1. **Herformuleer die Term:** Die term \(2^{1-2n}\) kan herskryf word as: \[ 2^{1-2n} = 2 \cdot 2^{-2n} = 2 \cdot \left(\frac{1}{4}\right)^n \] 2. **Identifiseer as 'n Geometriese Reeks:** Dus, die som word: \[ \sum_{n=1}^{\infty} 2^{1-2n} = 2 \sum_{n=1}^{\infty} \left(\frac{1}{4}\right)^n \] 3. **Gebruik die Formule vir 'n Geometriese Reeks:** Vir 'n geometriese reeks van die vorm \(\sum_{n=1}^{\infty} r^n\), is die som: \[ \sum_{n=1}^{\infty} r^n = \frac{r}{1 - r} \quad \text{vir} \ |r| < 1 \] Hier is \( r = \frac{1}{4} \), so: \[ \sum_{n=1}^{\infty} \left(\frac{1}{4}\right)^n = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} \] 4. **Bereken die Finale Som:** \[ 2 \cdot \frac{1}{3} = \frac{2}{3} \] **Dus, die waarde van die reeks is:** \[ \frac{2}{3} \]

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Om die waarde van die reeks \( \sum_{n=1}^{\infty} 2^{1-2 n} \) te bepaal, kan ons die terme van die reeks herskryf. Die terme kan herlei word as: \[ 2^{1-2n} = 2^{1} \cdot 2^{-2n} = 2 \cdot (2^{-2})^n \] Nou het ons die reeks: \[ \sum_{n=1}^{\infty} 2^{1-2n} = 2 \sum_{n=1}^{\infty} (2^{-2})^n = 2 \sum_{n=1}^{\infty} \left(\frac{1}{4}\right)^n \] Dit is 'n geometriese reeks met die eerste term \( a = \frac{1}{4} \) en 'n gemeenskaplike verhouding \( r = \frac{1}{4} \). Die som van 'n geometriese reeks kan bereken word volgens die formule: \[ S = \frac{a}{1 - r} \] Hier is \( a = \frac{1}{4} \) en \( r = \frac{1}{4} \). Dus, \[ \sum_{n=1}^{\infty} \left(\frac{1}{4}\right)^n = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} \] Nou kan ons die hele reeks bereken: \[ \sum_{n=1}^{\infty} 2^{1-2n} = 2 \cdot \frac{1}{3} = \frac{2}{3} \] Die waarde van die reeks \( \sum_{n=1}^{\infty} 2^{1-2 n} \) is dus \( \frac{2}{3} \).

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