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Math and Science => Technology & Calculator Open Topic
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polarBody

Newbie

Joined: 02 Sep 2009
Posts: 30

 Posted: 24 Sep 2009 12:43:59 am    Post subject: Does anyone know of a solution (in terms of n) to the following summation? ∑ [1/(i²+i)] where i goes from 1 to n I arrived at this summation as the solution to a recurrence equation, but I can go no further than this. Is this the end of the line? All I need is a gentle nudge in the right direction.Last edited by Guest on 05 Jul 2010 08:03:57 am; edited 1 time in total
Weregoose
Authentic INTJ

Super Elite (Last Title)

Joined: 25 Nov 2004
Posts: 3976

 Posted: 24 Sep 2009 02:07:49 am    Post subject: It helps if you look at each answer numerically. Convert each one to a fraction, and see what you can discover from that. Welcome to United-TI! Last edited by Guest on 24 Sep 2009 02:08:49 am; edited 1 time in total
polarBody

Newbie

Joined: 02 Sep 2009
Posts: 30

 Posted: 24 Sep 2009 10:38:42 am    Post subject: Thanks Weregoose. I tried that approach, but I feel like I'm in over my head. I was hoping that somebody would know of an already existing solution similar to this one: ∑ (i) = [n(n+1)]/2 where i goes from 1 to n I'm not trying to reinvent the wheel here if I don't have to. Apparently I need more than a gentle nudge. Last edited by Guest on 05 Jul 2010 08:04:17 am; edited 1 time in total
DarkerLine
ceci n'est pas une |

Super Elite (Last Title)

Joined: 04 Nov 2003
Posts: 8328

 Posted: 24 Sep 2009 01:00:15 pm    Post subject: If you try a few small values of n by hand, you'll see a pretty clear pattern. Maybe once you figure that out, and know what the answer ought to be in terms of n, you'll have a better idea of how to get it? If you're familiar with proofs by induction, this should be easy. Otherwise, it relies on you figuring out an alternate form in which to express 1/(i²+i).Last edited by Guest on 05 Jul 2010 08:04:48 am; edited 1 time in total
polarBody

Newbie

Joined: 02 Sep 2009
Posts: 30

 Posted: 24 Sep 2009 02:39:48 pm    Post subject: Ohh, duh! It's n/(n+1). Wow; I gave up way too easily on that one. Thanks for not giving away the answer, guys!
thornahawk
μολών λαβέ

Active Member

Joined: 27 Mar 2005
Posts: 569

 Posted: 28 Sep 2009 09:54:25 pm    Post subject: A late addition, but should you encounter more stuff like this, you'd do well to look into partial fraction decomposition. thornahawk
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