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

Lösung

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

\frac{1}{2}

Dezimale

0.5

Schritte zur Lösung

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

Wende die Regel der konstanten Multiplikation:

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

= 2 \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}

= 2 (n = 0 \sum \infty \frac{1}{5 ^{n}} - \frac{1}{5 ^{0}})

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

= 2 (\frac{5}{4} - \frac{1}{5 ^{0}})

Vereinfache

2 (\frac{5}{4} - \frac{1}{5 ^{0}}) : \frac{1}{2}

= \frac{1}{2}

(9 x + 1) y^{2} \frac{d y}{d x} + 2 x^{2} + 3 y^{3} = 0

d er i v a t i v e \frac{1}{x + 3}

y^{^{'}} = (x + y - 5)^{2}

\frac{d}{d x} (\frac{ln ( x )}{x + 2})

\frac{d}{d x} (x^{3} sin (5 x))