sum from n=1 to infinity of 3(5)^{-n}

Lösung

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

\frac{3}{4}

Dezimale

0.75

Schritte zur Lösung

n = 1 \sum \infty 3 \cdot 5^{- n}

Wende die Regel der konstanten Multiplikation:

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

= 3 \cdot n = 1 \sum \infty 5^{- n}

Vereinfache

5^{- n} : \frac{1}{5 ^{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}

x \to 0 lim (9 x - 8 x)

\int_{0}^{\frac{1}{4}} arcsin (2 x) d x

\frac{d}{d x} (\frac{5 x}{x ^{2} - 1})

d er i v a t i v e y = cos^{4} (x)

h \to 0 lim (3 x^{2} + 2 x h + h^{2})