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

Lösung

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

1

Schritte zur Lösung

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

Wende die Regel der konstanten Multiplikation:

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

= 2 \cdot 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}

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

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

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

Vereinfache

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

= 1

x \to 2 lim (3 + 2 x + x^{2})

x \to 3 lim (\frac{∣ x - 3 ∣}{x + 3})

\int sin (a) x d x

in v erse l a pl a ce \frac{1}{s ^{2} + 3 s + 2}

d er i v a t i v e f (t) = \frac{t ^{2}}{t ^{3} + 1}