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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    
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solution | Comment 1 of 3
let P(k)=S(k)S(k+1)

then we have
P(k)=S(k)*[S(k-1)-3/S(k)]
P(k)=S(k-1)S(k)-3
P(k)=P(k-1)-3
with P(0)=S(0)*S(1)=37*72=2664
thus
P(k)=2664-3k

now when P(k)=-3 then we have
-3=S(k)S(k-1)-3
S(k)S(k-1)=0
thus one of S(k) or S(k-1) is zero

so we have P(n)=-3
2664-3n=-3
3n=2667
n=889

so either S(888) or S(889) is zero

assume S(888)=0
then S(889)=S(887)-3/S(888)
which results in division by zero
thus S(889) is undefined
but if S(889) is undefined then so is P(889)=S(889)*S(890)
contradiction
thus S(888) is not zero and thus S(889) is zero

Thus n=889


  Posted by Daniel on 2016-01-28 09:23:03
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