What is the sum from k=1 to infinity of 90(0.1)^k ?

The sum from k=1 to infinity of 90(0.1)^k is 10

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from k=1 to infinity of 90(0.1)^k

k = 1 \sum \infty 90 (0.1)^{k}

10

Solution steps

k = 1 \sum \infty 90 \cdot 0. 1^{k}

Apply the constant multiplication rule:

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

= 90 \cdot k = 1 \sum \infty 0. 1^{k}

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

= 90 (k = 0 \sum \infty 0. 1^{k} - 0. 1^{0})

k = 0 \sum \infty 0. 1^{k} = 1.11111 \dots

= 90 (1.11111 \dots - 0. 1^{0})

Simplify

= 10

n = 0 \sum \infty \frac{4}{9 - n}

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

k = 1 \sum \infty 4 (- \frac{1}{5})^{4 k}

n = 0 \sum \infty n 2^{- n}

n = 1 \sum \infty 9 (0.4)^{n - 1}

What is the sum from k=1 to infinity of 90(0.1)^k ?
The sum from k=1 to infinity of 90(0.1)^k is 10