What is the sum from n=1 to infinity of (6/16)^n ?

The sum from n=1 to infinity of (6/16)^n is 3/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

sum from n=1 to infinity of (6/16)^n

n = 1 \sum \infty (\frac{6}{16})^{n}

\frac{3}{5}

Decimal

0.6

Solution steps

n = 1 \sum \infty (\frac{6}{16})^{n}

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

= n = 0 \sum \infty (\frac{6}{16})^{n} - (\frac{6}{16})^{0}

n = 0 \sum \infty (\frac{6}{16})^{n} = \frac{8}{5}

= \frac{8}{5} - (\frac{6}{16})^{0}

Simplify

\frac{8}{5} - (\frac{6}{16})^{0} : \frac{3}{5}

= \frac{3}{5}

What is the sum from n=1 to infinity of (6/16)^n ?
The sum from n=1 to infinity of (6/16)^n is 3/5