What is the sum from n=1 to infinity of (-3^n)/(8^n) ?

The sum from n=1 to infinity of (-3^n)/(8^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 (-3^n)/(8^n)

n = 1 \sum \infty \frac{- 3 ^{n}}{8 ^{n}}

- \frac{3}{5}

Decimal

- 0.6

Solution steps

n = 1 \sum \infty \frac{- 3 ^{n}}{8 ^{n}}

Apply the constant multiplication rule:

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

= - n = 1 \sum \infty \frac{3 ^{n}}{8 ^{n}}

Apply exponent rule:

\frac{a ^{c}}{b ^{c}} = (\frac{a}{b})^{c}

= - n = 1 \sum \infty (\frac{3}{8})^{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{3}{8})^{n} - (\frac{3}{8})^{0})

n = 0 \sum \infty (\frac{3}{8})^{n} = \frac{8}{5}

= - (\frac{8}{5} - (\frac{3}{8})^{0})

Simplify

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

= - \frac{3}{5}

d er i v a t i v e e^{2 x} - 3 x

3 e^{x} tan (y) d x + (2 - e^{x}) sec^{2} (y) d y = 0

t an g e n t y = 4 x^{3} - 4 x (- 1)

d er i v a t i v e ln (arccsc (x))

\frac{\partial}{\partial y} (x^{3} y^{5})

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