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The sum of series of 1/1*5+ 1/5*9 +1/9*13.......1/121*125 is

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The sum of series of 1/1*5+ 1/5*9 +1/9*13.......1/121*125 is [#permalink] New post 06 Nov 2009, 14:26
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The sum of series of 1/1*5+ 1/5*9 +1/9*13.......1/121*125 is

A. 28/221
B. 56/221
C. 56/225
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Re: Progression series [#permalink] New post 06 Nov 2009, 17:12
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virtualanimosity wrote:
Q.
The sum of series of
1/1*5+ 1/5*9 +1/9*13.......1/121*125 is

1.28/221
2.56/221
3.56/225


I wonder if there is an easier way but here is how I solved it.

N=\frac{1}{1*5}+\frac{1}{5*9} +\frac{1}{9*13}+.......+\frac{1}{121*125}

I really need 4 in nominator so let's multiply this by 4:

4N=\frac{4}{1*5}+ \frac{4}{5*9} +\frac{4}{9*13}+.......+\frac{4}{121*125}=\frac{(5-1)}{1*5}+\frac{(9-5)}{5*9} +\frac{(9-13)}{9*13}+.......+\frac{(125-121)}{121*125}=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{13}-\frac{1}{9}+...+\frac{1}{121}-\frac{1}{125}=1-\frac{1}{125}=\frac{124}{125}

4N=\frac{124}{125}

N=\frac{31}{125}

And after all that jazz no match with answer choices... But I can't see where I made a mistake.
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Re: Progression series [#permalink] New post 06 Nov 2009, 17:21
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I checked the answer choices given and noticed that #3 has a denominator 225 and nominator \frac{225-1}{4}=56, so if the last term were \frac{1}{221*225} instead of \frac{1}{121*125}, then the answer with my solution would be exactly \frac{56}{225}.

Maybe there is s typo?

BTW where are getting this questions? They are definitely not from GMAT.
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Re: Progression series [#permalink] New post 28 Sep 2012, 05:05
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Can we try this approach :

We have a series that appears as -

1/ 1 * 5 , 1/ 5 * 9 , 1 / 9 * 13 ...... 1 / 121 * 125


Can we split this into 2 series ??

First Series : 1/1 , 1/5, 1/9....1/121

Second Series 1/5/ 1/9, 1/13....1/125..
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Re: The sum of series of 1/1*5+ 1/5*9 +1/9*13.......1/121*125 is [#permalink] New post 05 Oct 2013, 10:14
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Re: The sum of series of 1/1*5+ 1/5*9 +1/9*13.......1/121*125 is   [#permalink] 05 Oct 2013, 10:14
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