What is the sum from n=0 to infinity of 3(-6/7)^n ?

The sum from n=0 to infinity of 3(-6/7)^n is 21/13

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sum from n=0 to infinity of 3(-6/7)^n

n = 0 \sum \infty 3 (- \frac{6}{7})^{n}

\frac{21}{13}

Solution steps

n = 0 \sum \infty 3 (- \frac{6}{7})^{n}

Apply the constant multiplication rule:

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

= 3 \cdot n = 0 \sum \infty (- \frac{6}{7})^{n}

Apply Series Geometric Test:

\frac{7}{13}

= 3 \cdot \frac{7}{13}

Simplify

3 \cdot \frac{7}{13} : \frac{21}{13}

= \frac{21}{13}

g (x) = \frac{7}{4} x^{2} - 3 x + 12

\int \frac{y}{y ^{2} + 1} d y

f (x) = \frac{x ^{2}}{( 9 + 4 x )}

\int \frac{4 x}{4 x ^{2} + 2 y ^{2}} d x

\int_{0}^{1} 3 e^{x} - 3 x e^{x^{2}} d x

What is the sum from n=0 to infinity of 3(-6/7)^n ?
The sum from n=0 to infinity of 3(-6/7)^n is 21/13