What is the sum from n=2 to infinity of (3/4)^n ?

The sum from n=2 to infinity of (3/4)^n is 9/4

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sum from n=2 to infinity of (3/4)^n

n = 2 \sum \infty (\frac{3}{4})^{n}

\frac{9}{4}

Solution steps

n = 2 \sum \infty (\frac{3}{4})^{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}{4})^{n} - (\frac{3}{4})^{0} - (\frac{3}{4})^{1}

n = 0 \sum \infty (\frac{3}{4})^{n} = 4

= 4 - (\frac{3}{4})^{0} - (\frac{3}{4})^{1}

Simplify

4 - (\frac{3}{4})^{0} - (\frac{3}{4})^{1} : \frac{9}{4}

= \frac{9}{4}

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What is the sum from n=2 to infinity of (3/4)^n ?
The sum from n=2 to infinity of (3/4)^n is 9/4