sum from n=1 to infinity of [53(1/2)^n]

Lösung

n = 1 \sum \infty [53 (\frac{1}{2})^{n}]

53

Schritte zur Lösung

n = 1 \sum \infty 53 (\frac{1}{2})^{n}

Wende die Regel der konstanten Multiplikation:

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

= 53 \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}

= 53 (n = 0 \sum \infty (\frac{1}{2})^{n} - (\frac{1}{2})^{0})

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

= 53 (2 - (\frac{1}{2})^{0})

Vereinfache

= 53

n = 0 \sum \infty \frac{8}{n ( n + 2 )}

n = 1 \sum \infty \frac{1}{n ( n - 1 )}

n = 1 \sum \infty \frac{1}{5 ^{n} - 1}

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

n = 1 \sum \infty (\frac{3 π}{4})^{n}