sum from n=1 to infinity of 23(2/5)^n

解答

n = 1 \sum \infty 23 (\frac{2}{5})^{n}

\frac{46}{3}

十进制

15.33333 \dots

求解步骤

n = 1 \sum \infty 23 (\frac{2}{5})^{n}

使用常数相乘法则:

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

= 23 \cdot n = 1 \sum \infty (\frac{2}{5})^{n}

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

= 23 (n = 0 \sum \infty (\frac{2}{5})^{n} - (\frac{2}{5})^{0})

n = 0 \sum \infty (\frac{2}{5})^{n} = \frac{5}{3}

= 23 (\frac{5}{3} - (\frac{2}{5})^{0})

化简

23 (\frac{5}{3} - (\frac{2}{5})^{0}) : \frac{46}{3}

= \frac{46}{3}

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

y (x + 3) + y^{^{'}} = 0

t an g e n t 8 + 2 x + x e^{x}

t an g e n t f (x) = \frac{4 x}{x + 3}, at x = - 5

x \to 1.7 lim (sin (x))