Payton Rasmussen

## Answered question

2022-09-06

Find nth term rule for $\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},...$

### Answer & Explanation

Kaleb Harrell

Beginner2022-09-07Added 14 answers

The series $\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},....$ can be written as
$\frac{1}{{2}^{1}},\frac{1}{{2}^{2}},\frac{1}{{2}^{3}},\frac{1}{{2}^{4}},....$
Hence ${n}^{th}$ term can be written as $\frac{1}{{2}^{n}}$.
Other way could be to treat it as a geometric series whose first term is ${a}_{1}$ and common ratio is r. The ${n}^{th}$ term of the series is then given by ${a}_{1}×{r}^{n-1}$.
As here ${a}_{1}=\frac{1}{2}$ and $r=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{4}×\frac{2}{1}=\frac{1}{2}$, the ${n}^{th}$ term is
$\frac{1}{2}×{\left(\frac{1}{2}\right)}^{n-1}$ or
$\frac{1}{2}×\frac{1}{{2}^{n-1}}=\frac{1}{{2}^{n}}$

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