What is the sum from n=0 to infinity of (400)(0.1)^n ?

The sum from n=0 to infinity of (400)(0.1)^n is 444.44444…

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=0 to infinity of (400)(0.1)^n

n = 0 \sum \infty (400) (0.1)^{n}

444.44444 \dots

Solution steps

n = 0 \sum \infty 400 \cdot 0. 1^{n}

Apply the constant multiplication rule:

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

= 400 \cdot n = 0 \sum \infty 0. 1^{n}

Apply Series Geometric Test:

1.11111 \dots

= 400 \cdot 1.11111 \dots

Simplify

= 444.44444 \dots

\frac{d y}{d x} = \frac{x ^{2} + 2}{3 y ^{2}}

(x^{2} + x)^{^{'}}

\frac{d}{d x} (\frac{( x ^{2} )}{sin ( x )})

y^{^{'}} - e^{6 x + y} = 0

\int x (x + 1)^{4} d x

What is the sum from n=0 to infinity of (400)(0.1)^n ?
The sum from n=0 to infinity of (400)(0.1)^n is 444.44444…