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Beliebt Rechnen >

sum from n=1 to infinity of (35)/(-6^n)

Lösung

n = 1 \sum \infty \frac{35}{- 6 ^{n}}

Lösung

- 7

Schritte zur Lösung

n = 1 \sum \infty \frac{35}{- 6 ^{n}}

Wende die Regel der konstanten Multiplikation:

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

= \frac{35}{- 1} \cdot n = 1 \sum \infty \frac{1}{6 ^{n}}

Vereinfache

= (- 35) \cdot n = 1 \sum \infty \frac{1}{6 ^{n}}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= (- 35) (n = 0 \sum \infty \frac{1}{6 ^{n}} - \frac{1}{6 ^{0}})

n = 0 \sum \infty \frac{1}{6 ^{n}} = \frac{6}{5}

= (- 35) (\frac{6}{5} - \frac{1}{6 ^{0}})

Vereinfache

(- 35) (\frac{6}{5} - \frac{1}{6 ^{0}}) : - 7

= - 7

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