sum from n=1 to infinity of-7(3/8)^n

Lösung

n = 1 \sum \infty - 7 (\frac{3}{8})^{n}

- \frac{21}{5}

Dezimale

- 4.2

Schritte zur Lösung

n = 1 \sum \infty - 7 (\frac{3}{8})^{n}

Wende die Regel der konstanten Multiplikation:

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

= - 7 \cdot n = 1 \sum \infty (\frac{3}{8})^{n}

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

= - 7 (n = 0 \sum \infty (\frac{3}{8})^{n} - (\frac{3}{8})^{0})

n = 0 \sum \infty (\frac{3}{8})^{n} = \frac{8}{5}

= - 7 (\frac{8}{5} - (\frac{3}{8})^{0})

Vereinfache

- 7 (\frac{8}{5} - (\frac{3}{8})^{0}) : - \frac{21}{5}

= - \frac{21}{5}

n = 1 \sum \infty \frac{1}{4 + 3 n}

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

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

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

n = 3 \sum \infty \frac{7}{3 ln ( n )}