sum from n=1 to infinity of 8*(0.2)^n

Lösung

n = 1 \sum \infty 8 \cdot (0.2)^{n}

2

Schritte zur Lösung

n = 1 \sum \infty 8 \cdot 0. 2^{n}

Wende die Regel der konstanten Multiplikation:

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

= 8 \cdot n = 1 \sum \infty 0. 2^{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 0. 2^{n} - 0. 2^{0})

n = 0 \sum \infty 0. 2^{n} = 1.25

= 8 (1.25 - 0. 2^{0})

Vereinfache

= 2

x \frac{d y}{d x} = y^{2} - y, y (2) = 1

t an g e n t f (x) = (10 x - 1)^{\frac{1}{2}}, at x = 1

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

\int \frac{e ^{x}}{e ^{2 x} + 8 e ^{x} + 16} d x

\int ln (x^{2} + 4) d x