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anonymous
 one year ago
Evaluate :
2014/2013 + 2014/2013 . 2012/2011 + 2014/2013 . 2012/2011 . 2010/2009 + .... + 2014/2013 . 2012/2011 . 2010/2009 . ... . 2/1  1
anonymous
 one year ago
Evaluate : 2014/2013 + 2014/2013 . 2012/2011 + 2014/2013 . 2012/2011 . 2010/2009 + .... + 2014/2013 . 2012/2011 . 2010/2009 . ... . 2/1  1

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amilapsn
 one year ago
Best ResponseYou've already chosen the best response.1Can you tell me what's the rth term of the series given? (beside 1)

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4Hint: try small (even) numbers instead of 2014, and deduce from there!

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.1i. e. : What's the rth term of this series? \[\frac{2014}{2013}+\frac{2014}{2013}\cdot\frac{2012}{2011}+\ldots+\frac{2014}{2013}\cdot\frac{2012}{2011}\ldots\frac{2}{1} \]

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4@amilapsn It's the sum that is interesting, individual terms are horrible.

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.1@mathmate yep but finding the rth term helps us to figure out a way to express it as f(r)f(r+1)

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.1It expands our understanding of the question too...

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.1If you REALLY don't want to find the rth term try to find a relationship between rth and (r+1)th term...

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4n=2 s(2)=2/11=1 n=4 s(4)=4/3+4/3*2/11 =4/3(1+2/1)1 =4/3(1+2)1 =41 =3 n=6 s(6)=6/5(1+4/3+4/3*2/1)1 =6/5(1+s(4)+1)1 =61 =5 So in general s(n)=n/(n1)(1+s(n2)+1) take it from here!

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4@amilapsn I missed the 1 because I was working without it in my mind, and intended to add it on later! Good catch! s(n)=n/(n1)(1+s(n2)+1) 1

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.1@mathmate by your method I don't see the need of a general S(n) term. can we can just say S(214)=2013?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4Yes, but mathematically we need to solve the recurrence relation, one way or another. If only the answer is required, yes, by all means! :)
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