sum from n=1 to infinity of (99)/(100^n)

Lösung

n = 1 \sum \infty \frac{99}{10 0 ^{n}}

1

Schritte zur Lösung

n = 1 \sum \infty \frac{99}{10 0 ^{n}}

Wende die Regel der konstanten Multiplikation:

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

= 99 \cdot n = 1 \sum \infty \frac{1}{10 0 ^{n}}

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

= 99 (n = 0 \sum \infty \frac{1}{10 0 ^{n}} - \frac{1}{10 0 ^{0}})

n = 0 \sum \infty \frac{1}{10 0 ^{n}} = \frac{100}{99}

= 99 (\frac{100}{99} - \frac{1}{10 0 ^{0}})

Vereinfache

99 (\frac{100}{99} - \frac{1}{10 0 ^{0}}) : 1

= 1

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

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

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

k = 1 \sum \infty k

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