How to do this question
- Thread starter mimie
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Hint: This is a Geometric series.
Please show us what you have tried and exactly where you are stuck.
Please follow the rules of posting in this forum, as enunciated at:
Please share your work/thoughts about this problem.
Otis
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It is not needed to be declared explicitly here. Three leading terms are given along with the expression for the last term.
Otis
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I do it like this
[MATH] \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^{n-1}}\\ a=\dfrac{1}{2},\ r=\dfrac{1}{2}\\ \text{using formula}\ S_{n-1}=\dfrac{a(1-r^{n-1})}{1-r}\\ S_{n-1}=\dfrac{\dfrac{1}{2}(1-(\dfrac{1}{2})^{n-1})}{1-\dfrac{1}{2}}\\ S_{n-1}=\dfrac{\dfrac{1}{2}-(\dfrac{1}{2})^{n}}{\dfrac{1}{2}}\\ S_{n-1}=1-(\dfrac{1}{2})^{n-1}\\ S_{n-1}=1-\dfrac{1}{2^{n-1}}\\ [/MATH]
[MATH] \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^{n-1}}\\ a=\dfrac{1}{2},\ r=\dfrac{1}{2}\\ \text{using formula}\ S_{n-1}=\dfrac{a(1-r^{n-1})}{1-r}\\ S_{n-1}=\dfrac{\dfrac{1}{2}(1-(\dfrac{1}{2})^{n-1})}{1-\dfrac{1}{2}}\\ S_{n-1}=\dfrac{\dfrac{1}{2}-(\dfrac{1}{2})^{n}}{\dfrac{1}{2}}\\ S_{n-1}=1-(\dfrac{1}{2})^{n-1}\\ S_{n-1}=1-\dfrac{1}{2^{n-1}}\\ [/MATH]
HallsofIvy
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Imagine that you want to go a distance of 1 km.
With one huge leap you jump 1/2 km. You have 1/2 km left to go.
Another leap takes you 1/4 km. So you have gone 1/2+ 1/4= 3/4 km and have 1/4 km left to go.
You are starting to tire so next you only jump 1/8 km the third time. You have gone 1/2+ 1/4+ 1/8= 4/8+ 2/8+ 1/8= 7/8 km and have 1/8 km left to go.
Next, your fourth, you jump 1/16 km. You have gone 1/2+ 1/4+ 1/8+ 1/16= 8/16+ 4/16+ 2/16+ 1/16= 15/16 km and have 1/16 km left to go.
Do you see that after having jumped "n" times you will still have \(\displaystyle 1/2^n\) km left to go? How long will it take you to realize that each time you jump, you are only going half way to the end so will never reach the end?!
With one huge leap you jump 1/2 km. You have 1/2 km left to go.
Another leap takes you 1/4 km. So you have gone 1/2+ 1/4= 3/4 km and have 1/4 km left to go.
You are starting to tire so next you only jump 1/8 km the third time. You have gone 1/2+ 1/4+ 1/8= 4/8+ 2/8+ 1/8= 7/8 km and have 1/8 km left to go.
Next, your fourth, you jump 1/16 km. You have gone 1/2+ 1/4+ 1/8+ 1/16= 8/16+ 4/16+ 2/16+ 1/16= 15/16 km and have 1/16 km left to go.
Do you see that after having jumped "n" times you will still have \(\displaystyle 1/2^n\) km left to go? How long will it take you to realize that each time you jump, you are only going half way to the end so will never reach the end?!
And yet Achilles can catch the tortoise.
HallsofIvy
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Achilles cheated!
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But Zeno couldn't .....
Achilles did not cheat. He simply appealed fron Zeno to Weierstrass; a dispute about the rules is not cheating. Anyway if it had been a fair race with everyone starting at the same starting line, the dispute would never have arisen.