What is the sum from n=1 to infinity of 2/25*(1/5)^n ?

The sum from n=1 to infinity of 2/25*(1/5)^n is 1/50

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sum from n=1 to infinity of 2/25*(1/5)^n

n = 1 \sum \infty \frac{2}{25} \cdot (\frac{1}{5})^{n}

\frac{1}{50}

Solution steps

n = 1 \sum \infty \frac{2}{25} (\frac{1}{5})^{n}

Apply the constant multiplication rule:

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

= \frac{2}{25} \cdot n = 1 \sum \infty (\frac{1}{5})^{n}

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

= \frac{2}{25} (n = 0 \sum \infty (\frac{1}{5})^{n} - (\frac{1}{5})^{0})

n = 0 \sum \infty (\frac{1}{5})^{n} = \frac{5}{4}

= \frac{2}{25} (\frac{5}{4} - (\frac{1}{5})^{0})

Simplify

\frac{2}{25} (\frac{5}{4} - (\frac{1}{5})^{0}) : \frac{1}{50}

= \frac{1}{50}

6 t e^{5 z}

4^{x}

y = \frac{e ^{x} + e ^{- x}}{1 + e ^{2 x}}

\int - x^{\frac{5}{2}} d x

\frac{d y}{d x} - 2 x y^{2} = 0

What is the sum from n=1 to infinity of 2/25*(1/5)^n ?
The sum from n=1 to infinity of 2/25*(1/5)^n is 1/50