ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある微分積分 >

sum from k=1 to infinity of 1/(2^k)

解

k = 1 \sum \infty \frac{1}{2 ^{k}}

解

1

解答ステップ

k = 1 \sum \infty \frac{1}{2 ^{k}}

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

= k = 0 \sum \infty \frac{1}{2 ^{k}} - \frac{1}{2 ^{0}}

k = 0 \sum \infty \frac{1}{2 ^{k}} = 2

= 2 - \frac{1}{2 ^{0}}

簡素化

= 1

人気の例

3x+y-2+(dy)/(dx)(x-1)=0

3 x + y - 2 + \frac{d y}{d x} (x - 1) = 0

sum from n=0 to infinity of 1/(2n+1)

n = 0 \sum \infty \frac{1}{2 n + 1}

integral of 1/(1+4x^2+4x^4)

\int \frac{1}{1 + 4 x ^{2} + 4 x ^{4}} d x

limit as x approaches pi/4 of 2sin(3x)

x \to \frac{π}{4} lim (2 sin (3 x))

derivative of (1-x^4^3)

\frac{d}{d x} ((1 - x^{4})^{3})