sum from n=1 to infinity of 30(0.1)^n

解答

n = 1 \sum \infty 30 (0.1)^{n}

3.33333 \dots

求解步骤

n = 1 \sum \infty 30 \cdot 0. 1^{n}

使用常数相乘法则:

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

= 30 \cdot n = 1 \sum \infty 0. 1^{n}

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

= 30 (n = 0 \sum \infty 0. 1^{n} - 0. 1^{0})

n = 0 \sum \infty 0. 1^{n} = 1.11111 \dots

= 30 (1.11111 \dots - 0. 1^{0})

化简

= 3.33333 \dots

n = 1 \sum \infty \frac{8}{n} + 3^{- n}

n = 1 \sum \infty \frac{1}{( 9 n ) ^{9}}

k = 2 \sum \infty (- 0.15)^{k}

n = 5 \sum \infty \frac{e ^{n}}{5 ^{n}}

n = 1 \sum \infty (\frac{9}{4})^{n}