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Beliebt Rechnen >

sum from k=0 to infinity of-5(3/2)^{k+2}

Lösung

k = 0 \sum \infty - 5 (\frac{3}{2})^{k + 2}

Lösung

divergiert

Schritte zur Lösung

k = 0 \sum \infty - 5 (\frac{3}{2})^{k + 2}

Wende die Regel der konstanten Multiplikation:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= - 5 \cdot k = 0 \sum \infty (\frac{3}{2})^{k + 2}

Vereinfache

(\frac{3}{2})^{k + 2} : (\frac{3}{2})^{k} (\frac{3}{2})^{2}

= - 5 \cdot k = 0 \sum \infty (\frac{3}{2})^{k} (\frac{3}{2})^{2}

Wende die Regel der konstanten Multiplikation:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= - 5 (\frac{3}{2})^{2} \cdot k = 0 \sum \infty (\frac{3}{2})^{k}

(\frac{3}{2})^{2} = \frac{9}{4}

= - 5 \cdot \frac{9}{4} \cdot k = 0 \sum \infty (\frac{3}{2})^{k}

Geometrischen Test der Reihe anwenden:

divergiert

= - 5 \cdot \frac{9}{4} divergiert

= divergiert

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