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 Last term zero (Posted on 2016-01-28)
{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.)
 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

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

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