sum from n=3 to infinity of 3/4*1/(4^n)

Lösung

n = 3 \sum \infty \frac{3}{4} \cdot \frac{1}{4 ^{n}}

\frac{1}{64}

Dezimale

0.015625

Schritte zur Lösung

n = 3 \sum \infty \frac{3}{4} \cdot \frac{1}{4 ^{n}}

Wende die Regel der konstanten Multiplikation:

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

= \frac{3}{4} \cdot n = 3 \sum \infty \frac{1}{4 ^{n}}

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

= \frac{3}{4} (n = 0 \sum \infty \frac{1}{4 ^{n}} - \frac{1}{4 ^{0}} - \frac{1}{4 ^{1}} - \frac{1}{4 ^{2}})

n = 0 \sum \infty \frac{1}{4 ^{n}} = \frac{4}{3}

= \frac{3}{4} (\frac{4}{3} - \frac{1}{4 ^{0}} - \frac{1}{4 ^{1}} - \frac{1}{4 ^{2}})

Vereinfache

\frac{3}{4} (\frac{4}{3} - \frac{1}{4 ^{0}} - \frac{1}{4 ^{1}} - \frac{1}{4 ^{2}}) : \frac{1}{64}

= \frac{1}{64}

\frac{\partial}{\partial z} (x z)

a re a - x^{2} + 3 x + 1, - x + 1

d er i v a t i v e t e^{- t}

x \to 2 lim (2 x^{2} - 4)

in v erse l a pl a ce \frac{s}{( s ^{2} - 4 ) ( s + 2 )}