sum from n=2 to infinity of 1/6 (5/6)^n

Lösung

n = 2 \sum \infty \frac{1}{6} (\frac{5}{6})^{n}

\frac{25}{36}

Dezimale

0.69444 \dots

Schritte zur Lösung

n = 2 \sum \infty \frac{1}{6} (\frac{5}{6})^{n}

Wende die Regel der konstanten Multiplikation:

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

= \frac{1}{6} \cdot n = 2 \sum \infty (\frac{5}{6})^{n}

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

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

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

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

Vereinfache

\frac{1}{6} (6 - (\frac{5}{6})^{0} - (\frac{5}{6})^{1}) : \frac{25}{36}

= \frac{25}{36}

k = 1 \sum \infty \frac{5}{( k + 4 ) ^{2}}

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

n = 1 \sum \infty \frac{1}{n ( 1 + n )}

n = 0 \sum \infty \frac{3}{n ^{2} + 3}

n = 0 \sum \infty \frac{3}{n ^{2} + 4}