What is the sum from n=1 to infinity of 3(-2/3)^n ?

The sum from n=1 to infinity of 3(-2/3)^n is -6/5

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sum from n=1 to infinity of 3(-2/3)^n

n = 1 \sum \infty 3 (- \frac{2}{3})^{n}

- \frac{6}{5}

Decimal

- 1.2

Solution steps

n = 1 \sum \infty 3 (- \frac{2}{3})^{n}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty (- \frac{2}{3})^{n}

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

= 3 (n = 0 \sum \infty (- \frac{2}{3})^{n} - (- \frac{2}{3})^{0})

n = 0 \sum \infty (- \frac{2}{3})^{n} = \frac{3}{5}

= 3 (\frac{3}{5} - (- \frac{2}{3})^{0})

Simplify

3 (\frac{3}{5} - (- \frac{2}{3})^{0}) : - \frac{6}{5}

= - \frac{6}{5}

What is the sum from n=1 to infinity of 3(-2/3)^n ?
The sum from n=1 to infinity of 3(-2/3)^n is -6/5