sum from n=2 to infinity of 5/(3^n)

Lösung

n = 2 \sum \infty \frac{5}{3 ^{n}}

\frac{5}{6}

Dezimale

0.83333 \dots

Schritte zur Lösung

n = 2 \sum \infty \frac{5}{3 ^{n}}

Wende die Regel der konstanten Multiplikation:

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

= 5 \cdot n = 2 \sum \infty \frac{1}{3 ^{n}}

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

= 5 (n = 0 \sum \infty \frac{1}{3 ^{n}} - \frac{1}{3 ^{0}} - \frac{1}{3 ^{1}})

n = 0 \sum \infty \frac{1}{3 ^{n}} = \frac{3}{2}

= 5 (\frac{3}{2} - \frac{1}{3 ^{0}} - \frac{1}{3 ^{1}})

Vereinfache

5 (\frac{3}{2} - \frac{1}{3 ^{0}} - \frac{1}{3 ^{1}}) : \frac{5}{6}

= \frac{5}{6}

d er i v a t i v e - \frac{32}{( t ) ^{3}} + \frac{4}{( t ) ^{2}}

\frac{\partial}{\partial x} (2 x y^{3} + 4)

\frac{d y}{d x} = \frac{y + 2 x y}{x}

e x t re m e f (x) = r^{2} - 10 x^{2}

a re a e^{- 2 x}, x \geq 3