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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

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  1. amilapsn
    • one year ago
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    Can you tell me what's the rth term of the series given? (beside -1)

  2. mathmate
    • one year ago
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    Hint: try small (even) numbers instead of 2014, and deduce from there!

  3. amilapsn
    • one year ago
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    i. e. : What's the rth term of this series? \[\frac{2014}{2013}+\frac{2014}{2013}\cdot\frac{2012}{2011}+\ldots+\frac{2014}{2013}\cdot\frac{2012}{2011}\ldots\frac{2}{1} \]

  4. mathmate
    • one year ago
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    @amilapsn It's the sum that is interesting, individual terms are horrible.

  5. amilapsn
    • one year ago
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    @mathmate yep but finding the rth term helps us to figure out a way to express it as f(r)-f(r+1)

  6. amilapsn
    • one year ago
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    It expands our understanding of the question too...

  7. amilapsn
    • one year ago
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    If you REALLY don't want to find the rth term try to find a relationship between rth and (r+1)th term...

  8. mathmate
    • one year ago
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    n=2 s(2)=2/1-1=1 n=4 s(4)=4/3+4/3*2/1-1 =4/3(1+2/1)-1 =4/3(1+2)-1 =4-1 =3 n=6 s(6)=6/5(1+4/3+4/3*2/1)-1 =6/5(1+s(4)+1)-1 =6-1 =5 So in general s(n)=n/(n-1)(1+s(n-2)+1) take it from here!

  9. amilapsn
    • one year ago
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    ha ha that's clever!

  10. mathmate
    • one year ago
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    :)

  11. mathmate
    • one year ago
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    @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/(n-1)(1+s(n-2)+1) -1

  12. amilapsn
    • one year ago
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    @mathmate by your method I don't see the need of a general S(n) term. can we can just say S(2-14)=2013?

  13. amilapsn
    • one year ago
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    *S(2014)

  14. mathmate
    • one year ago
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    Yes, but mathematically we need to solve the recurrence relation, one way or another. If only the answer is required, yes, by all means! :)

  15. amilapsn
    • one year ago
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    Nice!

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