All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers > Sequences
Last term zero (Posted on 2016-01-28) Difficulty: 3 of 5
{S(0), S(1), S(2), ....., S(n) } is a sequence of real numbers such that:
S(0) = 37, S(1) = 72, ....., S(n) = 0, and:

S(k+1) = S(k-1) – 3/S(k) for k=0,1,2,...., n-1

Find n

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Spoiler | Comment 2 of 3 |
The problem is not possible.

Suppose there is a term n such that S(n)=0
That means 0 = S(n-2) - 3/S(n-1)
so S(n-1)*S(n-2)=3

But S(n-1) = S(n-3) - 3/S(n-2)
and substituting in gives
[S(n-3) - 3/S(n-2)]*S(n-2)=3
S(n-3)*S(n-2)=0  <--- Algebra error.  This should be 6.
Now S(n-2) cannot be 0, so S(n-3)=0

This means S(n-1)=0-3/S(n-2)
meaning S(n-2)*S(n-1)=-3
Which is a contradiction.

So there is no n such that S(n)=0.

***** Edit *****
Daniel's solution seems valid and a quick check with Excel seems to confirm this.  We posted at nearly the same time so I didn't see his until I hit enter.  I think something must be wrong with the above.  I wonder what.

Now that I've got S(n-3)*S(n-2)=6 I could probably show S(n-(a+1))*S(n-a)=3a and work backwards to give the same solution as Daniel.

Edited on January 28, 2016, 10:46 am
  Posted by Jer on 2016-01-28 09:34:50

Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information