What is the sum from n=0 to infinity of 3*(-2/5)^n ?

The sum from n=0 to infinity of 3*(-2/5)^n is 15/7

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sum from n=0 to infinity of 3*(-2/5)^n

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

\frac{15}{7}

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Geometric Test:

\frac{5}{7}

= 3 \cdot \frac{5}{7}

Simplify

3 \cdot \frac{5}{7} : \frac{15}{7}

= \frac{15}{7}

f (x) = \frac{2 x}{x - 1}, at x = 2

\int (x^{2} + 1) e^{2 x} d x

f (x) = 5 e^{x}

\frac{d}{d x} (\frac{3}{4} x^{\frac{3}{2}})

\frac{1}{( 3 x ) ^{2}}

What is the sum from n=0 to infinity of 3*(-2/5)^n ?
The sum from n=0 to infinity of 3*(-2/5)^n is 15/7