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

Lösung

n = 1 \sum \infty 3 \cdot (\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}

n \to \infty lim (n^{2} e^{- 3 n})

in v erse l a pl a ce \frac{( s - 3 )}{( s ^{2} + 4 s + 13 )}

y^{^{'}} - 7 y = - x

\int (5 x + 4)^{5} d x

\frac{d}{d x} ((3 x^{4} - 1) (x - 1))