What is the sum from n=3 to infinity of 1/3 (2/5)^n ?

The sum from n=3 to infinity of 1/3 (2/5)^n is 8/225

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=3 to infinity of 1/3 (2/5)^n

n = 3 \sum \infty \frac{1}{3} (\frac{2}{5})^{n}

\frac{8}{225}

Solution steps

n = 3 \sum \infty \frac{1}{3} (\frac{2}{5})^{n}

Apply the constant multiplication rule:

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

= \frac{1}{3} \cdot n = 3 \sum \infty (\frac{2}{5})^{n}

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

= \frac{1}{3} (n = 0 \sum \infty (\frac{2}{5})^{n} - (\frac{2}{5})^{0} - (\frac{2}{5})^{1} - (\frac{2}{5})^{2})

n = 0 \sum \infty (\frac{2}{5})^{n} = \frac{5}{3}

= \frac{1}{3} (\frac{5}{3} - (\frac{2}{5})^{0} - (\frac{2}{5})^{1} - (\frac{2}{5})^{2})

Simplify

\frac{1}{3} (\frac{5}{3} - (\frac{2}{5})^{0} - (\frac{2}{5})^{1} - (\frac{2}{5})^{2}) : \frac{8}{225}

= \frac{8}{225}

\int \frac{1}{5 ^{4 x}} d x

\frac{d}{d x} (\frac{3 cos ( x )}{3 x ^{2} + 3 x - 1})

- 3 y^{^{'}} + \frac{2 y}{x} = \frac{( y ^{4} )}{x ^{4}}

\frac{d y}{d x} = \frac{x - y - 3}{x - 3}

\frac{1}{0.012 s ^{2} + 1.55 s + 50}

What is the sum from n=3 to infinity of 1/3 (2/5)^n ?
The sum from n=3 to infinity of 1/3 (2/5)^n is 8/225