sum from n=1 to infinity of (4/10)^n*3

Lösung

n = 1 \sum \infty (\frac{4}{10})^{n} \cdot 3

2

Schritte zur Lösung

n = 1 \sum \infty (\frac{4}{10})^{n} \cdot 3

Wende die Regel der konstanten Multiplikation:

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

= 3 \cdot n = 1 \sum \infty (\frac{4}{10})^{n}

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

= 3 (n = 0 \sum \infty (\frac{4}{10})^{n} - (\frac{4}{10})^{0})

n = 0 \sum \infty (\frac{4}{10})^{n} = \frac{5}{3}

= 3 (\frac{5}{3} - (\frac{4}{10})^{0})

Vereinfache

3 (\frac{5}{3} - (\frac{4}{10})^{0}) : 2

= 2

\frac{d}{d x} (ln (cosh (8 x)))

\frac{d ^{2}}{d x ^{2}} (x^{2} e^{- 2 x})

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t an g e n t \frac{1}{16} tan^{2} (x) + cos (x)

\int \frac{x ^{2} + 10 x - 1}{x ^{3} - x} d x