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

The sum from n=2 to infinity of 2/(5^n) is 1/10

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

n = 2 \sum \infty \frac{2}{5 ^{n}}

\frac{1}{10}

Solution steps

n = 2 \sum \infty \frac{2}{5 ^{n}}

Apply the constant multiplication rule:

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

= 2 \cdot n = 2 \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}

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

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

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

Simplify

2 (\frac{5}{4} - \frac{1}{5 ^{0}} - \frac{1}{5 ^{1}}) : \frac{1}{10}

= \frac{1}{10}

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y^{^{'}} - 4 y = e^{- 4 t}

\frac{\partial}{\partial t} (\frac{s - r}{t})

What is the sum from n=2 to infinity of 2/(5^n) ?
The sum from n=2 to infinity of 2/(5^n) is 1/10