sum from n=2 to infinity of-1/(4^n)

Lösung

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

- \frac{1}{12}

Dezimale

- 0.08333 \dots

Schritte zur Lösung

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

Wende die Regel der konstanten Multiplikation:

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

= - n = 2 \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}

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

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

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

Vereinfache

- (\frac{4}{3} - \frac{1}{4 ^{0}} - \frac{1}{4 ^{1}}) : - \frac{1}{12}

= - \frac{1}{12}

x \to 0 lim (\frac{9 ^{x} - 5 ^{x}}{x})

\frac{\partial}{\partial y} (\frac{y ln ( y )}{x - 1})

\int \frac{2}{x 9 + x ^{2}} d x

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d er i v a t i v e f (x) = \frac{( 1 - 4 x )}{3 + x}