sum from n=1 to infinity of 3(1/5)^n

Lösung

n = 1 \sum \infty 3 (\frac{1}{5})^{n}

\frac{3}{4}

Dezimale

0.75

Schritte zur Lösung

n = 1 \sum \infty 3 (\frac{1}{5})^{n}

Wende die Regel der konstanten Multiplikation:

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

= 3 \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}

= 3 (n = 0 \sum \infty (\frac{1}{5})^{n} - (\frac{1}{5})^{0})

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

= 3 (\frac{5}{4} - (\frac{1}{5})^{0})

Vereinfache

3 (\frac{5}{4} - (\frac{1}{5})^{0}) : \frac{3}{4}

= \frac{3}{4}

k = 1 \sum \infty \frac{k ^{k}}{2 ^{k}}

i = 1 \sum \infty 3 e^{i}

n = 1 \sum \infty - \frac{1}{1 8 ^{n}}

n = 1 \sum \infty (cos (14))^{n}

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