sum from n=2 to infinity of (1/3)^n

Lösung

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

\frac{1}{6}

Dezimale

0.16666 \dots

Schritte zur Lösung

n = 2 \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} - (\frac{1}{3})^{1}

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

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

Vereinfache

\frac{3}{2} - (\frac{1}{3})^{0} - (\frac{1}{3})^{1} : \frac{1}{6}

= \frac{1}{6}

\frac{d}{d x} (5 x^{4} - 2 x^{3} - 3 x + 2)

y^{^{''}} + 8 y^{^{'}} + 7 y = 0

\int sin (x) cos (x) 1 - sin^{4} (x) d x

\frac{\partial}{\partial x} ((3 y) \cdot (2 x)^{- 1})

p a r i t y f (x) = sin (tan (2 x))