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

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

n = 1 \sum \infty - \frac{3}{2 ^{n}}

- 3

Solution steps

n = 1 \sum \infty - \frac{3}{2 ^{n}}

Apply the constant multiplication rule:

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

= - 3 \cdot n = 1 \sum \infty \frac{1}{2 ^{n}}

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

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

n = 0 \sum \infty \frac{1}{2 ^{n}} = 2

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

Simplify

= - 3

\int \frac{x ^{2}}{13 - 4 x + x ^{2}} d x

\frac{- 3 x - 16 x}{2 x ^{\frac{1}{3}}}

y^{^{''}} + 20 y^{^{'}} + 200 y = 0

y^{^{''}} + 8 y^{^{'}} + 13 y = 0

x \frac{d y}{d x} - y = x^{2} y^{2}

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