What is the sum from k=0 to infinity of 9*(3/5)^k ?

The sum from k=0 to infinity of 9*(3/5)^k is 45/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from k=0 to infinity of 9*(3/5)^k

k = 0 \sum \infty 9 \cdot (\frac{3}{5})^{k}

\frac{45}{2}

Solution steps

k = 0 \sum \infty 9 (\frac{3}{5})^{k}

Apply the constant multiplication rule:

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

= 9 \cdot k = 0 \sum \infty (\frac{3}{5})^{k}

Apply Series Geometric Test:

\frac{5}{2}

= 9 \cdot \frac{5}{2}

Simplify

9 \cdot \frac{5}{2} : \frac{45}{2}

= \frac{45}{2}

n = 2 \sum \infty (0.2)^{n}

n = 0 \sum \infty cos (4 n)

n = 0 \sum \infty e^{- 10 n}

k = 1 \sum \infty \frac{2 ^{k}}{( k ) !}

n = 1 \sum \infty (- 1)^{n + 1} \cdot n

What is the sum from k=0 to infinity of 9*(3/5)^k ?
The sum from k=0 to infinity of 9*(3/5)^k is 45/2