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# Annoying analysis question watch

1. It may be shown that the sum from k=1 to infinity = lim n to infinity (r + r2 + r3 + ... rn ) = r/(1-r) for any lrl<1. Use this fact to write the following recurring decimal in rational form p/q where p and q are whole numbers.

0.161616161616....

Thanks
2. (Original post by *Sarah...*)
It may be shown that the sum from k=1 to infinity = lim n to infinity (r + r2 + r3 + ... rn ) = r/(1-r) for any lrl<1. Use this fact to write the following recurring decimal in rational form p/q where p and q are whole numbers.

0.161616161616....

Thanks
I'm not sure what method they are talking about but it can be done like this
Let x = 0.16161616...*****(1)
Then 100x = 16.161616****(2)
(2) - (1)
99x =16
x=16/99
3. (Original post by steve2005)
I'm not sure what method they are talking about but it can be done like this
Let x = 0.16161616...*****(1)
Then 100x = 16.161616****(2)
(2) - (1)
99x =16
x=16/99
oh thanks for that, I didn't realise it was so simple
4. Oh thanks for that i didn't realise it was so simple!
5. (Original post by *Sarah...*)
Oh thanks for that i didn't realise it was so simple!
Yes, steve's fraction is correct BUT the question says use a particular method... I don't think steve has used this method so ' no marks'

I don't know the method either. We need Gaz or AlphaNumeric to put us on the right path.

Edit: I've used the required method.

0.1616161616 = 16/100 + 16/10000 + 16/1000000....
This is a GP with r= 1/100
r/(1-r) = [16/100]/[1- 1/100]=[16/100]/[99/100] =16/99
6. (Original post by carol05)
Yes, steve's fraction is correct BUT the question says use a particular method... I don't think steve has used this method so ' no marks'

I don't know the method either. We need Gaz or AlphaNumeric to put us on the right path.

Edit: I've used the required method.

0.1616161616 = 16/100 + 16/10000 + 16/1000000....
This is a GP with r= 1/100
r/(1-r) = [16/100]/[1- 1/100]=[16/100]/[99/100] =16/99
Yes, I think this fulfills the requirements of the question. I always prefer it when the question says. Hence, or otherwise...... this then allows you to use any method, although the one suggested is often the easiest..

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Updated: November 18, 2005
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