sum from n=1 to infinity of (1/8)^n

解答

n = 1 \sum \infty (\frac{1}{8})^{n}

\frac{1}{7}

十进制

0.14285 \dots

求解步骤

n = 1 \sum \infty (\frac{1}{8})^{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{1}{8})^{n} - (\frac{1}{8})^{0}

n = 0 \sum \infty (\frac{1}{8})^{n} = \frac{8}{7}

= \frac{8}{7} - (\frac{1}{8})^{0}

化简

\frac{8}{7} - (\frac{1}{8})^{0} : \frac{1}{7}

= \frac{1}{7}

\frac{d}{d x} (\frac{x + 3}{x ^{3} + x - 2})

\int \frac{1}{( 8 x - 1 ) ^{3}} d x

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

d er i v a t i v e f (x) = a^{2}

d er i v a t i v e \frac{x ^{2} - 49}{x - 7}