What is the sum from n=0 to infinity of 7/(5^n) ?

The sum from n=0 to infinity of 7/(5^n) is 35/4

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/(5^n)

n = 0 \sum \infty \frac{7}{5 ^{n}}

\frac{35}{4}

Solution steps

n = 0 \sum \infty \frac{7}{5 ^{n}}

Apply the constant multiplication rule:

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

= 7 \cdot n = 0 \sum \infty \frac{1}{5 ^{n}}

Apply Series Geometric Test:

\frac{5}{4}

= 7 \cdot \frac{5}{4}

Simplify

7 \cdot \frac{5}{4} : \frac{35}{4}

= \frac{35}{4}

\int - 7 sec (2 x) tan (2 x) d x

\int \frac{3 x}{x ^{2} + 1} d x

\int \frac{2 x - 3}{x ^{2} - 3 x + 2} d x

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

\int cot^{3} (5 x) d x

What is the sum from n=0 to infinity of 7/(5^n) ?
The sum from n=0 to infinity of 7/(5^n) is 35/4