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

解

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

\frac{64}{147}

十進法表記

0.43537 \dots

解答ステップ

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}

簡素化

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

= \frac{64}{147}