sum from n=1 to infinity of 4*(4/7)^n

解答

n = 1 \sum \infty 4 \cdot (\frac{4}{7})^{n}

\frac{16}{3}

十进制

5.33333 \dots

求解步骤

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

使用常数相乘法则:

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

= 4 \cdot n = 1 \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}

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

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

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

化简

4 (\frac{7}{3} - (\frac{4}{7})^{0}) : \frac{16}{3}

= \frac{16}{3}

\int (\frac{x ( x - 8 )}{x + 1})^{2} d x

d er i v a t i v e f (x) = 5 x (sin (x) + cos (x))

\int sin (π + 2 y) - sin (2 y) d y

\frac{\partial}{\partial y} (x y e^{y} + e^{y} sin (x))

\frac{\partial}{\partial x} (16 - x^{2} - y^{2})