sum from n=3 to infinity of (4/7)^n

Lösung

n = 3 \sum \infty (\frac{4}{7})^{n}

\frac{64}{147}

Dezimale

0.43537 \dots

Schritte zur Lösung

n = 3 \sum \infty (\frac{4}{7})^{n}

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

= n = 0 \sum \infty (\frac{4}{7})^{n} - (\frac{4}{7})^{0} - (\frac{4}{7})^{1} - (\frac{4}{7})^{2}

n = 0 \sum \infty (\frac{4}{7})^{n} = \frac{7}{3}

= \frac{7}{3} - (\frac{4}{7})^{0} - (\frac{4}{7})^{1} - (\frac{4}{7})^{2}

Vereinfache

\frac{7}{3} - (\frac{4}{7})^{0} - (\frac{4}{7})^{1} - (\frac{4}{7})^{2} : \frac{64}{147}

= \frac{64}{147}

\int \frac{1}{cos ^{2} ( 4 x )} d x

\frac{d}{d x} (e^{x} - 4)

\int \frac{x ^{2} + 2 x + 5}{( x + 1 ) ( x ^{2} + 1 )} d x

\int_{0}^{1} \frac{12}{x ^{2} - 4} d x

a re a \frac{64}{16 + x ^{2}}, 1