sum from n=1 to infinity of 5^{-n}

Lösung

n = 1 \sum \infty 5^{- n}

\frac{1}{4}

Dezimale

0.25

Schritte zur Lösung

n = 1 \sum \infty 5^{- n}

Vereinfache

5^{- n} : \frac{1}{5 ^{n}}

= n = 1 \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}

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

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

= \frac{5}{4} - \frac{1}{5 ^{0}}

Vereinfache

\frac{5}{4} - \frac{1}{5 ^{0}} : \frac{1}{4}

= \frac{1}{4}

\frac{\partial}{\partial x} (arcsin (\frac{l}{x}))

n \to \infty lim (\frac{n}{2})

d er i v a t i v e 0.09 e^{x} x

n = 0 \sum \infty 8 (\frac{1}{2})^{n - 1}

roo t s x^{2} y