What is the sum from n=0 to infinity of 7/(3^n) ?

The sum from n=0 to infinity of 7/(3^n) is 21/2

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 7/(3^n)

n = 0 \sum \infty \frac{7}{3 ^{n}}

\frac{21}{2}

Solution steps

n = 0 \sum \infty \frac{7}{3 ^{n}}

Apply the constant multiplication rule:

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

= 7 \cdot n = 0 \sum \infty \frac{1}{3 ^{n}}

Apply Series Geometric Test:

\frac{3}{2}

= 7 \cdot \frac{3}{2}

Simplify

7 \cdot \frac{3}{2} : \frac{21}{2}

= \frac{21}{2}

\frac{d}{d x} (cosh (k x))

\int \frac{1}{tan ^{2} ( u ) + 1} d u

x \to 1 - lim (\frac{1}{x ^{2} - 1})

\int \frac{2}{3 x} d x

x \to \frac{π}{2} lim (csc (x))

What is the sum from n=0 to infinity of 7/(3^n) ?
The sum from n=0 to infinity of 7/(3^n) is 21/2