What is the sum from n=1 to infinity of 1/(5^{n+3)} ?

The sum from n=1 to infinity of 1/(5^{n+3)} is 1/500

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sum from n=1 to infinity of 1/(5^{n+3)}

n = 1 \sum \infty \frac{1}{5 ^{n + 3}}

\frac{1}{500}

Solution steps

n = 1 \sum \infty \frac{1}{5 ^{n + 3}}

Simplify

5^{n + 3} : 5^{n} \cdot 5^{3}

= n = 1 \sum \infty \frac{1}{5 ^{n} \cdot 5 ^{3}}

Apply the constant multiplication rule:

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

= \frac{1}{5 ^{3}} \cdot n = 1 \sum \infty \frac{1}{5 ^{n}}

Simplify

= \frac{1}{125} \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{1}{125} (n = 0 \sum \infty \frac{1}{5 ^{n}} - \frac{1}{5 ^{0}})

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

= \frac{1}{125} (\frac{5}{4} - \frac{1}{5 ^{0}})

Simplify

\frac{1}{125} (\frac{5}{4} - \frac{1}{5 ^{0}}) : \frac{1}{500}

= \frac{1}{500}

\int \frac{3 x ^{2}}{125} d x

sin (cos (tan (x)))

\frac{d}{d x} (1 + cos (2 x))

ln (x y^{9}) = x y

What is the sum from n=1 to infinity of 1/(5^{n+3)} ?
The sum from n=1 to infinity of 1/(5^{n+3)} is 1/500