sum from n=1 to infinity of-15/2*(2/3)^n

Lösung

n = 1 \sum \infty - \frac{15}{2} \cdot (\frac{2}{3})^{n}

- 15

Schritte zur Lösung

n = 1 \sum \infty - \frac{15}{2} (\frac{2}{3})^{n}

Wende die Regel der konstanten Multiplikation:

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

= - \frac{15}{2} \cdot n = 1 \sum \infty (\frac{2}{3})^{n}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= - \frac{15}{2} (n = 0 \sum \infty (\frac{2}{3})^{n} - (\frac{2}{3})^{0})

n = 0 \sum \infty (\frac{2}{3})^{n} = 3

= - \frac{15}{2} (3 - (\frac{2}{3})^{0})

Vereinfache

- \frac{15}{2} (3 - (\frac{2}{3})^{0}) : - 15

= - 15

in v erse l a pl a ce \frac{s ^{2}}{s ^{2} + 1}

\frac{\partial}{\partial x} (3 x^{3} - 9 x + 9 x y^{2})

\int_{1}^{2} \frac{e ^{\frac{1}{x ^{3}}}}{x ^{4}} d x

n or ma l y = (3 x^{2} + 1)^{2}, (0, 1)

\frac{\partial}{\partial x} (x^{2} - y^{2} - z^{2})