What is the sum from k=0 to infinity of 3(-1/4)^k ?

The sum from k=0 to infinity of 3(-1/4)^k is 12/5

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sum from k=0 to infinity of 3(-1/4)^k

k = 0 \sum \infty 3 (- \frac{1}{4})^{k}

\frac{12}{5}

Solution steps

k = 0 \sum \infty 3 (- \frac{1}{4})^{k}

Apply the constant multiplication rule:

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

= 3 \cdot k = 0 \sum \infty (- \frac{1}{4})^{k}

Apply Series Geometric Test:

\frac{4}{5}

= 3 \cdot \frac{4}{5}

Simplify

3 \cdot \frac{4}{5} : \frac{12}{5}

= \frac{12}{5}

n = 1 \sum \infty \frac{3}{5 n + 1}

i = 1 \sum \infty 7 + 3 i

n = 1 \sum \infty (- 0.45)^{n}

k = 1 \sum \infty \frac{k}{k + 1}

n = 0 \sum \infty (\frac{3}{π})^{n}

What is the sum from k=0 to infinity of 3(-1/4)^k ?
The sum from k=0 to infinity of 3(-1/4)^k is 12/5