What is the sum from n=2 to infinity of (3^n)/(10^n) ?

The sum from n=2 to infinity of (3^n)/(10^n) is 9/70

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sum from n=2 to infinity of (3^n)/(10^n)

n = 2 \sum \infty \frac{3 ^{n}}{1 0 ^{n}}

\frac{9}{70}

Solution steps

n = 2 \sum \infty \frac{3 ^{n}}{1 0 ^{n}}

Apply exponent rule:

\frac{a ^{c}}{b ^{c}} = (\frac{a}{b})^{c}

= n = 2 \sum \infty (\frac{3}{10})^{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{3}{10})^{n} - (\frac{3}{10})^{0} - (\frac{3}{10})^{1}

n = 0 \sum \infty (\frac{3}{10})^{n} = \frac{10}{7}

= \frac{10}{7} - (\frac{3}{10})^{0} - (\frac{3}{10})^{1}

Simplify

\frac{10}{7} - (\frac{3}{10})^{0} - (\frac{3}{10})^{1} : \frac{9}{70}

= \frac{9}{70}

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What is the sum from n=2 to infinity of (3^n)/(10^n) ?
The sum from n=2 to infinity of (3^n)/(10^n) is 9/70