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Beliebt Rechnen >

sum from n=2 to infinity of (3^n)/(12^n)

Lösung

n = 2 \sum \infty \frac{3 ^{n}}{1 2 ^{n}}

Lösung

\frac{1}{12}

+1

Dezimale

0.08333 \dots

Schritte zur Lösung

n = 2 \sum \infty \frac{3 ^{n}}{1 2 ^{n}}

Vereinfache

\frac{3 ^{n}}{1 2 ^{n}} : \frac{1}{4 ^{n}}

= n = 2 \sum \infty \frac{1}{4 ^{n}}

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

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

n = 0 \sum \infty \frac{1}{4 ^{n}} = \frac{4}{3}

= \frac{4}{3} - \frac{1}{4 ^{0}} - \frac{1}{4 ^{1}}

Vereinfache

\frac{4}{3} - \frac{1}{4 ^{0}} - \frac{1}{4 ^{1}} : \frac{1}{12}

= \frac{1}{12}

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