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

Lösung

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

\frac{1}{2}

Dezimale

0.5

Schritte zur Lösung

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

Vereinfache

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

n = 0 \sum \infty \frac{1}{3 ^{n}} = \frac{3}{2}

= \frac{3}{2} - \frac{1}{3 ^{0}}

Vereinfache

\frac{3}{2} - \frac{1}{3 ^{0}} : \frac{1}{2}

= \frac{1}{2}

in v erse l a pl a ce \frac{s}{2 s + 1}

\frac{d y}{d x} + \frac{3}{x} y = 16 y^{3}, y (1) = 1

\int_{- 2}^{2} 2 - 4 - x^{2} d x

in v erse l a pl a ce \frac{1}{s + 5}

t an g e n t f (x) = - 5 x^{2}, at x = 8