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

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

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sum from n=1 to infinity of 2/(6^n)

n = 1 \sum \infty \frac{2}{6 ^{n}}

\frac{2}{5}

Solution steps

n = 1 \sum \infty \frac{2}{6 ^{n}}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty \frac{1}{6 ^{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}{6 ^{n}} - \frac{1}{6 ^{0}})

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

= 2 (\frac{6}{5} - \frac{1}{6 ^{0}})

Simplify

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

= \frac{2}{5}

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

k = 3 \sum \infty \frac{1}{( k - 2 ) ^{5}}

n = 3 \sum \infty \frac{1}{n !}

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

n = 0 \sum \infty \frac{n + 1}{2 n + 1}

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