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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{5}{2}

Solution steps

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

Apply the constant multiplication rule:

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

= 5 \cdot n = 2 \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}} - \frac{1}{2 ^{1}})

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

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

Simplify

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

= \frac{5}{2}

t e^{- 5 t}

5 t ln (t)

\int_{0}^{b} \frac{1}{16 - x ^{2}} d x

\frac{s}{s ^{2} + 2 s + 1}

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