sum from n=1 to infinity of 60(0.1)^n

Lösung

n = 1 \sum \infty 60 (0.1)^{n}

6.66666 \dots

Schritte zur Lösung

n = 1 \sum \infty 60 \cdot 0. 1^{n}

Wende die Regel der konstanten Multiplikation:

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

= 60 \cdot n = 1 \sum \infty 0. 1^{n}

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

= 60 (n = 0 \sum \infty 0. 1^{n} - 0. 1^{0})

n = 0 \sum \infty 0. 1^{n} = 1.11111 \dots

= 60 (1.11111 \dots - 0. 1^{0})

Vereinfache

= 6.66666 \dots

n = 0 \sum \infty (\frac{5}{9})^{n}

n = 1 \sum \infty \frac{n !}{9 4 ^{n}}

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

n = 1 \sum \infty \frac{1 ^{n}}{4 ^{n}}

x = 0 \sum \infty x^{2}