sum from n=1 to infinity of 6/(4n^2+4n)

Lösung

n = 1 \sum \infty \frac{6}{4 n ^{2} + 4 n}

\frac{3}{2}

Dezimale

1.5

Schritte zur Lösung

n = 1 \sum \infty \frac{6}{4 n ^{2} + 4 n}

Wende die Regel der konstanten Multiplikation:

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

= 6 \cdot n = 1 \sum \infty \frac{1}{4 n ^{2} + 4 n}

Teleskopreihen-Test anwenden:

\frac{1}{4}

= 6 \cdot \frac{1}{4}

Vereinfache

6 \cdot \frac{1}{4} : \frac{3}{2}

= \frac{3}{2}

k = 9 \sum \infty \frac{7}{k ^{6.6}}

n = 1 \sum \infty \frac{n}{n + 11}

n = 1 \sum \infty (\frac{13}{2})^{n}

k = 1 \sum \infty \frac{4 k}{e ^{k}}

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