sum from n=1 to infinity of 8(1/5)^n

Lösung

n = 1 \sum \infty 8 (\frac{1}{5})^{n}

2

Schritte zur Lösung

n = 1 \sum \infty 8 (\frac{1}{5})^{n}

Wende die Regel der konstanten Multiplikation:

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

= 8 \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}

= 8 (n = 0 \sum \infty (\frac{1}{5})^{n} - (\frac{1}{5})^{0})

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

= 8 (\frac{5}{4} - (\frac{1}{5})^{0})

Vereinfache

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

= 2

\frac{d}{d x} ((5 x)^{7 x})

x \to 0 + lim (\frac{5 x}{sin ( x )})

x \to 3 - lim (- x^{2} + \frac{5}{2} x + 3)

\frac{d}{d x} (\frac{5 x ^{6} + 4 x ^{5} + 3 x ^{4}}{6 x ^{2}})

d er i v a t i v e \frac{x ^{2} - 2 e ^{x}}{x + 3 e ^{x}}