What is the sum from n=1 to infinity of (1/5)^{n+2} ?

The sum from n=1 to infinity of (1/5)^{n+2} is 1/100

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sum from n=1 to infinity of (1/5)^{n+2}

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

\frac{1}{100}

Decimal

0.01

Solution steps

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

Simplify

(\frac{1}{5})^{n + 2} : (\frac{1}{5})^{n} (\frac{1}{5})^{2}

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

Apply the constant multiplication rule:

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

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

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

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

Simplify

\frac{1}{25} (\frac{5}{4} - (\frac{1}{5})^{0}) : \frac{1}{100}

= \frac{1}{100}

What is the sum from n=1 to infinity of (1/5)^{n+2} ?
The sum from n=1 to infinity of (1/5)^{n+2} is 1/100