sum from k=2 to infinity of 4/(7^k)

Lösung

k = 2 \sum \infty \frac{4}{7 ^{k}}

\frac{2}{21}

Dezimale

0.09523 \dots

Schritte zur Lösung

k = 2 \sum \infty \frac{4}{7 ^{k}}

Wende die Regel der konstanten Multiplikation:

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

= 4 \cdot k = 2 \sum \infty \frac{1}{7 ^{k}}

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

= 4 (k = 0 \sum \infty \frac{1}{7 ^{k}} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}})

k = 0 \sum \infty \frac{1}{7 ^{k}} = \frac{7}{6}

= 4 (\frac{7}{6} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}})

Vereinfache

4 (\frac{7}{6} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}}) : \frac{2}{21}

= \frac{2}{21}

k = 1 \sum \infty (1)^{k}

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

n = 1 \sum \infty \frac{4}{7 n - 1}

n = 9 \sum \infty \frac{4}{8 ^{n}}

n = 0 \sum \infty \frac{n ^{2}}{1}