What is the sum from n=0 to infinity of 1800(1/3)^n ?

The sum from n=0 to infinity of 1800(1/3)^n is 2700

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sum from n=0 to infinity of 1800(1/3)^n

n = 0 \sum \infty 1800 (\frac{1}{3})^{n}

2700

Solution steps

n = 0 \sum \infty 1800 (\frac{1}{3})^{n}

Apply the constant multiplication rule:

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

= 1800 \cdot n = 0 \sum \infty (\frac{1}{3})^{n}

Apply Series Geometric Test:

\frac{3}{2}

= 1800 \cdot \frac{3}{2}

Simplify

1800 \cdot \frac{3}{2} : 2700

= 2700

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What is the sum from n=0 to infinity of 1800(1/3)^n ?
The sum from n=0 to infinity of 1800(1/3)^n is 2700