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

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

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 5/(2^n)

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

5

Solution steps

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

Apply the constant multiplication rule:

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

= 5 \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}

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

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

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

Simplify

= 5

y = \frac{2 x}{x + 1}, at x = 1

2 \frac{d y}{d x} + 4 y = 1

f (x) = cos (x^{4})

y = t

x \to 0 lim (sin^{5} (2 x^{2}))

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