sum from k=1 to infinity of 4^{-k}

Lösung

k = 1 \sum \infty 4^{- k}

\frac{1}{3}

Dezimale

0.33333 \dots

Schritte zur Lösung

k = 1 \sum \infty 4^{- k}

Vereinfache

4^{- k} : \frac{1}{4 ^{k}}

= k = 1 \sum \infty \frac{1}{4 ^{k}}

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

= k = 0 \sum \infty \frac{1}{4 ^{k}} - \frac{1}{4 ^{0}}

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

= \frac{4}{3} - \frac{1}{4 ^{0}}

Vereinfache

\frac{4}{3} - \frac{1}{4 ^{0}} : \frac{1}{3}

= \frac{1}{3}

k = 0 \sum \infty \frac{1}{900 + 3 k}

n = 1 \sum \infty 6^{- 2 n}

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

y^{^{'}} = 3 x y + 5 y

\frac{d y}{d x} = \frac{3 x ^{2} + 4 x + 2}{2 ( y - 1 )}