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

The sum from n=1 to infinity of 3(5)^{-n} is 3/4

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

n = 1 \sum \infty 3 (5)^{- n}

\frac{3}{4}

Decimal

0.75

Solution steps

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

Apply the constant multiplication rule:

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

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

Simplify

5^{- n} : \frac{1}{5 ^{n}}

= 3 \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}

= 3 (n = 0 \sum \infty \frac{1}{5 ^{n}} - \frac{1}{5 ^{0}})

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

= 3 (\frac{5}{4} - \frac{1}{5 ^{0}})

Simplify

3 (\frac{5}{4} - \frac{1}{5 ^{0}}) : \frac{3}{4}

= \frac{3}{4}

What is the sum from n=1 to infinity of 3(5)^{-n} ?
The sum from n=1 to infinity of 3(5)^{-n} is 3/4