What is the sum from n=1 to infinity of 2*0.2^n ?

The sum from n=1 to infinity of 2*0.2^n is 0.5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=1 to infinity of 2*0.2^n

n = 1 \sum \infty 2 \cdot 0. 2^{n}

0.5

Solution steps

n = 1 \sum \infty 2 \cdot 0. 2^{n}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty 0. 2^{n}

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

= 2 (n = 0 \sum \infty 0. 2^{n} - 0. 2^{0})

n = 0 \sum \infty 0. 2^{n} = 1.25

= 2 (1.25 - 0. 2^{0})

Simplify

= 0.5

3 x^{2} e^{y} + (x^{3} e^{y} - 1) y^{^{'}} = 0

\int_{1}^{8} \frac{3}{x} d x

\int \frac{( x ^{2} )}{( 15 + 6 x - 9 x ^{2} ) ^{\frac{3}{2}}} d x

\frac{d}{d x} (e^{x^{\frac{1}{2}}})

f (x) = 2 x^{3} - 3 x^{2} + 3, at x = 1

What is the sum from n=1 to infinity of 2*0.2^n ?
The sum from n=1 to infinity of 2*0.2^n is 0.5