What is the sum from n=1 to infinity of 4^{-n} ?

The sum from n=1 to infinity of 4^{-n} is 1/3

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 4^{-n}

n = 1 \sum \infty 4^{- n}

\frac{1}{3}

Solution steps

n = 1 \sum \infty 4^{- n}

Simplify

4^{- n} : \frac{1}{4 ^{n}}

= n = 1 \sum \infty \frac{1}{4 ^{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{1}{4 ^{n}} - \frac{1}{4 ^{0}}

n = 0 \sum \infty \frac{1}{4 ^{n}} = \frac{4}{3}

= \frac{4}{3} - \frac{1}{4 ^{0}}

Simplify

\frac{4}{3} - \frac{1}{4 ^{0}} : \frac{1}{3}

= \frac{1}{3}

\int \frac{5 - 50 x}{64 - 25 x ^{2}} d x

x \to 4 + lim (e^{\frac{7}{( 4 - x )}})

4 sin (x) cos (x)

- x^{2} + 7 x, x^{2} - 3 x

\int - x cos (n x) d x

What is the sum from n=1 to infinity of 4^{-n} ?
The sum from n=1 to infinity of 4^{-n} is 1/3