What is the sum from n=1 to infinity of 6/(5^n) ?

The sum from n=1 to infinity of 6/(5^n) is 3/2

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sum from n=1 to infinity of 6/(5^n)

n = 1 \sum \infty \frac{6}{5 ^{n}}

\frac{3}{2}

Solution steps

n = 1 \sum \infty \frac{6}{5 ^{n}}

Apply the constant multiplication rule:

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

= 6 \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}

= 6 (n = 0 \sum \infty \frac{1}{5 ^{n}} - \frac{1}{5 ^{0}})

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

= 6 (\frac{5}{4} - \frac{1}{5 ^{0}})

Simplify

6 (\frac{5}{4} - \frac{1}{5 ^{0}}) : \frac{3}{2}

= \frac{3}{2}

n = 0 \sum \infty \frac{2 ^{n}}{n ^{4}}

n = 1 \sum \infty (0.7)^{n}

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

k = 0 \sum \infty (\frac{1}{19})^{k}

n = 1 \sum \infty (\frac{3}{2})^{3 - 2 n}

What is the sum from n=1 to infinity of 6/(5^n) ?
The sum from n=1 to infinity of 6/(5^n) is 3/2