What is the sum from n=2 to infinity of 6/(3^{2n)} ?

The sum from n=2 to infinity of 6/(3^{2n)} is 1/12

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sum from n=2 to infinity of 6/(3^{2n)}

n = 2 \sum \infty \frac{6}{3 ^{2 n}}

\frac{1}{12}

Solution steps

n = 2 \sum \infty \frac{6}{3 ^{2 n}}

Apply the constant multiplication rule:

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

= 6 \cdot n = 2 \sum \infty \frac{1}{3 ^{2 n}}

3^{2 n} = 9^{n}

= 6 \cdot n = 2 \sum \infty \frac{1}{9 ^{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}{9 ^{n}} - \frac{1}{9 ^{0}} - \frac{1}{9 ^{1}})

n = 0 \sum \infty \frac{1}{9 ^{n}} = \frac{9}{8}

= 6 (\frac{9}{8} - \frac{1}{9 ^{0}} - \frac{1}{9 ^{1}})

Simplify

6 (\frac{9}{8} - \frac{1}{9 ^{0}} - \frac{1}{9 ^{1}}) : \frac{1}{12}

= \frac{1}{12}

n = 0 \sum \infty \frac{ln ( n ^{2} )}{n}

n = 0 \sum \infty \frac{3}{3 + n ^{2}}

n = 1 \sum \infty \frac{n}{n ^{2} + 8}

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

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

What is the sum from n=2 to infinity of 6/(3^{2n)} ?
The sum from n=2 to infinity of 6/(3^{2n)} is 1/12