sum from n=2 to infinity of 6/(5^n)

Lösung

n = 2 \sum \infty \frac{6}{5 ^{n}}

\frac{3}{10}

Dezimale

0.3

Schritte zur Lösung

n = 2 \sum \infty \frac{6}{5 ^{n}}

Wende die Regel der konstanten Multiplikation:

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

= 6 \cdot n = 2 \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}

= 6 (n = 0 \sum \infty \frac{1}{5 ^{n}} - \frac{1}{5 ^{0}} - \frac{1}{5 ^{1}})

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

= 6 (\frac{5}{4} - \frac{1}{5 ^{0}} - \frac{1}{5 ^{1}})

Vereinfache

6 (\frac{5}{4} - \frac{1}{5 ^{0}} - \frac{1}{5 ^{1}}) : \frac{3}{10}

= \frac{3}{10}

\int cos^{4} (x s i) n^{3} x d x

\int_{1}^{4} \frac{u - 2}{u} d u

\frac{\partial}{\partial y} ((x^{2} y - y^{3})^{5})

\frac{d y}{d x} = \frac{cos ( 9 x )}{e ^{9 y}}

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