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Sequence. (1 Viewer)

Fus Ro Dah

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Sorry, I made a typo. The first term should read T_{n+1}.
 

barbernator

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don't we need the value of T0 or something to evaluate T2012?
 

bleakarcher

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Don't we need some initial condition in order to find T_2012 like the value of T_0?
 

Fus Ro Dah

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My apologies, everybody. I am not thinking clearly at the moment. The initial value T_0 = T.
 

barbernator

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<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{T}{1@plus;4044121T}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{T}{1+4044121T}" title="\frac{T}{1+4044121T}" /></a> edited.

?
 
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RealiseNothing

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On second attempt, and this time not adding an extra 'T' which screwed me up before I get:



Where denotes the k'th triangular number.
 

seanieg89

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Falls apart from the substitution S_n=1/T_n.

In the case T=1 this yields: T_n= 2/(n^2-n+2).
 

RealiseNothing

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Since triangular numbers can be worked out using we get:



Therefore:

 
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deswa1

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subbed in twice and observed pattern... lol. Is it correct?
Haha I do this for so many questions. Not sure if you're right- haven't looked at the question. I think this question is the only one out of the set of like 4 that Fus just posted that I'll have a chance at solving lol
 

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