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

解答

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

\frac{8}{75}

十进制

0.10666 \dots

求解步骤

n = 3 \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}

= n = 0 \sum \infty (\frac{4}{10})^{n} - (\frac{4}{10})^{0} - (\frac{4}{10})^{1} - (\frac{4}{10})^{2}

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

= \frac{5}{3} - (\frac{4}{10})^{0} - (\frac{4}{10})^{1} - (\frac{4}{10})^{2}

化简

\frac{5}{3} - (\frac{4}{10})^{0} - (\frac{4}{10})^{1} - (\frac{4}{10})^{2} : \frac{8}{75}

= \frac{8}{75}

\frac{\partial}{\partial x} (x^{2} y + 10 x y)

d er i v a t i v e y = ln (8) x^{2}

d er i v a t i v e (2 x - 9)^{2}

\int x^{3} e^{x^{2} - 1} d x

t an g e n t y = x, (16, 4)