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# Help!

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625
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+531

Find the sum of the base-2 geometric series $$0.1_2-0.01_2+0.001_2-0.0001_2+0.00001_2\ldots$$; give your answer as a fraction in which the numerator and denominator are both expressed in base 10.

This is urgent!

Dec 20, 2018

#2
+24947
+11

Find the sum of the base-2 geometric series
$$0.1_2-0.01_2+0.001_2-0.0001_2+0.00001_2\ldots$$;
give your answer as a fraction in which the numerator and denominator are both expressed in base 10.

$$\begin{array}{|rcll|} \hline 0.1_{2} = 1\cdot 2^{-1} &=& \frac{1}{2} \\ 0.01_{2} = 1\cdot 2^{-2} &=& \frac{1}{4} \\ 0.001_{2} = 1\cdot 2^{-3} &=& \frac{1}{8} \\ 0.0001_{2} = 1\cdot 2^{-4} &=& \frac{1}{16} \\ 0.00001_{2} = 1\cdot 2^{-5} &=& \frac{1}{32} \\ \ldots \\ \hline \end{array}$$

Infinite Geometric Series:

$$\begin{array}{|rcll|} \hline s &=& \dfrac{1}{2} - \dfrac{1}{4} + \dfrac{1}{8} - \dfrac{1}{16} + \dfrac{1}{32} \pm \ldots \quad | \quad a=\frac{1}{2},~ r = -\frac{1}{2} \\\\ s &=& \dfrac{a}{1-r} \quad | \quad |r| < 1. \\\\ s &=& \dfrac{\frac{1}{2}}{1- \left(-\frac{1}{2} \right)} \\\\ s &=& \dfrac{\frac{1}{2}}{1+\frac{1}{2}} \\\\ s &=& \dfrac{\frac{1}{2}}{\frac{3}{2}} \\\\ s &=& \dfrac{1}{2} \cdot \dfrac{2}{3} \\\\ \mathbf{s} & \mathbf{=} & \mathbf{\dfrac{1}{3}} \\ \hline \end{array}$$

Dec 20, 2018
edited by heureka  Dec 20, 2018
edited by heureka  Dec 20, 2018

#1
+6185
+1

xxxx

.
Dec 20, 2018
edited by Rom  Dec 20, 2018
#2
+24947
+11

Find the sum of the base-2 geometric series
$$0.1_2-0.01_2+0.001_2-0.0001_2+0.00001_2\ldots$$;
give your answer as a fraction in which the numerator and denominator are both expressed in base 10.

$$\begin{array}{|rcll|} \hline 0.1_{2} = 1\cdot 2^{-1} &=& \frac{1}{2} \\ 0.01_{2} = 1\cdot 2^{-2} &=& \frac{1}{4} \\ 0.001_{2} = 1\cdot 2^{-3} &=& \frac{1}{8} \\ 0.0001_{2} = 1\cdot 2^{-4} &=& \frac{1}{16} \\ 0.00001_{2} = 1\cdot 2^{-5} &=& \frac{1}{32} \\ \ldots \\ \hline \end{array}$$

Infinite Geometric Series:

$$\begin{array}{|rcll|} \hline s &=& \dfrac{1}{2} - \dfrac{1}{4} + \dfrac{1}{8} - \dfrac{1}{16} + \dfrac{1}{32} \pm \ldots \quad | \quad a=\frac{1}{2},~ r = -\frac{1}{2} \\\\ s &=& \dfrac{a}{1-r} \quad | \quad |r| < 1. \\\\ s &=& \dfrac{\frac{1}{2}}{1- \left(-\frac{1}{2} \right)} \\\\ s &=& \dfrac{\frac{1}{2}}{1+\frac{1}{2}} \\\\ s &=& \dfrac{\frac{1}{2}}{\frac{3}{2}} \\\\ s &=& \dfrac{1}{2} \cdot \dfrac{2}{3} \\\\ \mathbf{s} & \mathbf{=} & \mathbf{\dfrac{1}{3}} \\ \hline \end{array}$$

heureka Dec 20, 2018
edited by heureka  Dec 20, 2018
edited by heureka  Dec 20, 2018
#3
+6185
+1

why did you switch it to an alternating series?

Rom  Dec 20, 2018
edited by Rom  Dec 20, 2018
edited by Rom  Dec 20, 2018
#4
+24947
+9

Hello Rom,

i switch it to an alternating series, because the question is: 0.1-  0.012 + 0.0012 - 0.00012 + 0.000012 +- ...

and the answer is a fraction.

$$\begin{array}{|rcll|} \hline r &=& \dfrac{ {\color{red}-}\dfrac{1}{4} } { \dfrac{1}{2} } = {\color{red}-}\dfrac{1}{2} \\\\ r &=& \dfrac{ \dfrac{1}{8} } { {\color{red}-}\dfrac{1}{4} } = {\color{red}-}\dfrac{1}{2} \\ \ldots \\ \hline \end{array}$$

heureka  Dec 20, 2018
edited by heureka  Dec 20, 2018
#5
+6185
+1

d**n you're right....

time for new glasses.

Rom  Dec 20, 2018
#6
+111321
0

LOL!!!!....I know the feeling.....!!!

CPhill  Dec 20, 2018