sum from n=1 to infinity of (30)/(3^n)

解答

n = 1 \sum \infty \frac{30}{3 ^{n}}

15

求解步骤

n = 1 \sum \infty \frac{30}{3 ^{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 \frac{1}{3 ^{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 \frac{1}{3 ^{n}} - \frac{1}{3 ^{0}})

n = 0 \sum \infty \frac{1}{3 ^{n}} = \frac{3}{2}

= 30 (\frac{3}{2} - \frac{1}{3 ^{0}})

化简

30 (\frac{3}{2} - \frac{1}{3 ^{0}}) : 15

= 15

a re a y = \frac{2}{1 + x ^{2}}, y = ∣ x ∣

\int \frac{2 x}{4 - x ^{2}} d x

\int 4 e^{7 x} d x

\frac{d y}{d x} = x^{4} y^{- 7}

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