sum from n=1 to infinity of (50)/(5^n)

Lösung

n = 1 \sum \infty \frac{50}{5 ^{n}}

\frac{25}{2}

Dezimale

12.5

Schritte zur Lösung

n = 1 \sum \infty \frac{50}{5 ^{n}}

Wende die Regel der konstanten Multiplikation:

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

= 50 \cdot n = 1 \sum \infty \frac{1}{5 ^{n}}

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

= 50 (n = 0 \sum \infty \frac{1}{5 ^{n}} - \frac{1}{5 ^{0}})

n = 0 \sum \infty \frac{1}{5 ^{n}} = \frac{5}{4}

= 50 (\frac{5}{4} - \frac{1}{5 ^{0}})

Vereinfache

50 (\frac{5}{4} - \frac{1}{5 ^{0}}) : \frac{25}{2}

= \frac{25}{2}

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