sum from n=2 to infinity of 4/(7^n)

Lösung

n = 2 \sum \infty \frac{4}{7 ^{n}}

\frac{2}{21}

Dezimale

0.09523 \dots

Schritte zur Lösung

n = 2 \sum \infty \frac{4}{7 ^{n}}

Wende die Regel der konstanten Multiplikation:

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

= 4 \cdot n = 2 \sum \infty \frac{1}{7 ^{n}}

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

= 4 (n = 0 \sum \infty \frac{1}{7 ^{n}} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}})

n = 0 \sum \infty \frac{1}{7 ^{n}} = \frac{7}{6}

= 4 (\frac{7}{6} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}})

Vereinfache

4 (\frac{7}{6} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}}) : \frac{2}{21}

= \frac{2}{21}

d er i v a t i v e f (x) = a^{5} + cos^{5} (x)

t an g e n t f (x) = x^{2} - 3 x, at x = 2

t an g e n t y = \frac{6 x}{x ^{2} + 1}, (1, 1)

\frac{d}{d x} (x^{3} - 2 x + 7)

\frac{d}{d x} (cos (ln (x)) - sin (ln (x)))