sum from n=2 to infinity of 8/(4^{2n)}

Lösung

n = 2 \sum \infty \frac{8}{4 ^{2 n}}

\frac{1}{30}

Dezimale

0.03333 \dots

Schritte zur Lösung

n = 2 \sum \infty \frac{8}{4 ^{2 n}}

Wende die Regel der konstanten Multiplikation:

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

= 8 \cdot n = 2 \sum \infty \frac{1}{4 ^{2 n}}

4^{2 n} = 1 6^{n}

= 8 \cdot n = 2 \sum \infty \frac{1}{1 6 ^{n}}

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

= 8 (n = 0 \sum \infty \frac{1}{1 6 ^{n}} - \frac{1}{1 6 ^{0}} - \frac{1}{1 6 ^{1}})

n = 0 \sum \infty \frac{1}{1 6 ^{n}} = \frac{16}{15}

= 8 (\frac{16}{15} - \frac{1}{1 6 ^{0}} - \frac{1}{1 6 ^{1}})

Vereinfache

8 (\frac{16}{15} - \frac{1}{1 6 ^{0}} - \frac{1}{1 6 ^{1}}) : \frac{1}{30}

= \frac{1}{30}

y^{^{'}} = \frac{- 2 x y}{1 + x ^{2}}

\frac{\partial}{\partial x} (42 y^{8} x^{5} - 30 y^{7} x^{4})

d er i v a t i v e cos (2 arctan (\frac{x}{2}))

\int \frac{u ^{2}}{u ^{2} - 8 u} d u

\frac{d}{d x} (x^{\frac{1}{3}} (x + 4))