What is the sum from n=1 to infinity of 1/(2^{3n)} ?

The sum from n=1 to infinity of 1/(2^{3n)} is 1/7

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sum from n=1 to infinity of 1/(2^{3n)}

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

\frac{1}{7}

Solution steps

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

2^{3 n} = 8^{n}

= n = 1 \sum \infty \frac{1}{8 ^{n}}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= n = 0 \sum \infty \frac{1}{8 ^{n}} - \frac{1}{8 ^{0}}

n = 0 \sum \infty \frac{1}{8 ^{n}} = \frac{8}{7}

= \frac{8}{7} - \frac{1}{8 ^{0}}

Simplify

\frac{8}{7} - \frac{1}{8 ^{0}} : \frac{1}{7}

= \frac{1}{7}

f (x) = \frac{1}{2 x + 3}, at x = 2

\frac{d}{d x} (y cos (x))

cos (3 x^{2} + 4) \cdot 6 x

\int \frac{1}{3 x ^{2} + 6 x + 6} d x

sin (x), x = 0, y = 0, x = \frac{π}{6}

What is the sum from n=1 to infinity of 1/(2^{3n)} ?
The sum from n=1 to infinity of 1/(2^{3n)} is 1/7