Payton Rasmussen

2022-09-06

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

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 $a}_{1}\times {r}^{n-1$.

As here $a}_{1}=\frac{1}{2$ and $r=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{4}\times \frac{2}{1}=\frac{1}{2}$, the $n}^{th$ term is

$\frac{1}{2}\times {\left(\frac{1}{2}\right)}^{n-1}$ or

$\frac{1}{2}\times \frac{1}{{2}^{n-1}}=\frac{1}{{2}^{n}}$

$\frac{1}{{2}^{1}},\frac{1}{{2}^{2}},\frac{1}{{2}^{3}},\frac{1}{{2}^{4}},....$

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 $a}_{1}\times {r}^{n-1$.

As here $a}_{1}=\frac{1}{2$ and $r=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{4}\times \frac{2}{1}=\frac{1}{2}$, the $n}^{th$ term is

$\frac{1}{2}\times {\left(\frac{1}{2}\right)}^{n-1}$ or

$\frac{1}{2}\times \frac{1}{{2}^{n-1}}=\frac{1}{{2}^{n}}$

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