Question

Evaluate :
2014/2013 + 2014/2013 . 2012/2011 + 2014/2013 . 2012/2011 . 2010/2009 + .... + 2014/2013 . 2012/2011 . 2010/2009 . ... . 2/1 - 1

Answer 1

Can you tell me what's the rth term of the series given? (beside -1)

Answer 2

Hint:
try small (even) numbers instead of 2014, and deduce from there!

Answer 3

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

Answer 4

@amilapsn
It's the sum that is interesting, individual terms are horrible.

Answer 5

@mathmate yep but finding the rth term helps us to figure out a way to express it as f(r)-f(r+1)

Answer 6

It expands our understanding of the question too...

Answer 7

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

Answer 8

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!

Answer 9

ha ha that's clever!

Answer 10

:)

Answer 11

@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

Answer 12

@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?

Answer 13

*S(2014)

Answer 14

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

Answer 15

Nice!