sum from n=1 to infinity of 7/(2^n)

Lösung

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

7

Schritte zur Lösung

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

Wende die Regel der konstanten Multiplikation:

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

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

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

= 7 (n = 0 \sum \infty \frac{1}{2 ^{n}} - \frac{1}{2 ^{0}})

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

= 7 (2 - \frac{1}{2 ^{0}})

Vereinfache

= 7

\frac{d}{d x} (\frac{9 x ^{2}}{x ^{2} + 2})

\int cos (x) cos (n x) d x

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

\int \frac{1}{x ^{2} + 8 x + 52} d x

x^{2} y^{^{'}} = y^{2} + 3 x y