sum from j=1 to infinity of 3(1/4)^j

Lösung

j = 1 \sum \infty 3 (\frac{1}{4})^{j}

1

Schritte zur Lösung

j = 1 \sum \infty 3 (\frac{1}{4})^{j}

Wende die Regel der konstanten Multiplikation:

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

= 3 \cdot j = 1 \sum \infty (\frac{1}{4})^{j}

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

= 3 (j = 0 \sum \infty (\frac{1}{4})^{j} - (\frac{1}{4})^{0})

j = 0 \sum \infty (\frac{1}{4})^{j} = \frac{4}{3}

= 3 (\frac{4}{3} - (\frac{1}{4})^{0})

Vereinfache

3 (\frac{4}{3} - (\frac{1}{4})^{0}) : 1

= 1

n = 1 \sum \infty \frac{n}{n + 8}

n = 9 \sum \infty \frac{1}{9 n ^{4} + 8}

n = 0 \sum \infty (0. 2^{n}) 0.8

k = 7 \sum \infty \frac{1}{k ^{2} - 5 k + 6}

k = 0 \sum \infty (\frac{4}{7})^{k}