What is the sum from n=0 to infinity of-6*(-1/3)^n ?

The sum from n=0 to infinity of-6*(-1/3)^n is -9/2

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sum from n=0 to infinity of-6*(-1/3)^n

n = 0 \sum \infty - 6 \cdot (- \frac{1}{3})^{n}

- \frac{9}{2}

Solution steps

n = 0 \sum \infty - 6 (- \frac{1}{3})^{n}

Apply the constant multiplication rule:

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

= - 6 \cdot n = 0 \sum \infty (- \frac{1}{3})^{n}

Apply Series Geometric Test:

\frac{3}{4}

= - 6 \cdot \frac{3}{4}

Simplify

- 6 \cdot \frac{3}{4} : - \frac{9}{2}

= - \frac{9}{2}

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What is the sum from n=0 to infinity of-6*(-1/3)^n ?
The sum from n=0 to infinity of-6*(-1/3)^n is -9/2