sum from n=2 to infinity of 6/(7^n)

解答

n = 2 \sum \infty \frac{6}{7 ^{n}}

\frac{1}{7}

十进制

0.14285 \dots

求解步骤

n = 2 \sum \infty \frac{6}{7 ^{n}}

使用常数相乘法则:

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

= 6 \cdot n = 2 \sum \infty \frac{1}{7 ^{n}}

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

= 6 (n = 0 \sum \infty \frac{1}{7 ^{n}} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}})

n = 0 \sum \infty \frac{1}{7 ^{n}} = \frac{7}{6}

= 6 (\frac{7}{6} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}})

化简

6 (\frac{7}{6} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}}) : \frac{1}{7}

= \frac{1}{7}

x \to - 3 lim (\frac{2 x}{x ^{2} + 4})

x y^{^{'}} - 2 y = 2 x^{2}

y^{^{''}} + y^{^{'}} - 42 y = 0

\int \frac{5 x ^{2}}{( a ^{2} - x ^{2} ) ^{\frac{3}{2}}} d x

t an g e n t f (x) = e^{x}, at x = ln (10)