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Ti-1 + 2*(Ti-1)-1 = 2*Ti + (Ti)-1 (Posted on 2009-09-20) Difficulty: 3 of 5
{T0, T1, T2,..., T1995} is a sequence of positive real numbers with T0 = T1995, that satisfies this set of equations:

Ti-1 + 2*(Ti-1)-1 = 2*Ti + (Ti)-1, for all i=1, 2,...., 1995

Determine the maximum value of T0.

No Solution Yet Submitted by K Sengupta    
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a start Comment 1 of 1

ok so we have
let T(k)=x and T(k-1)=y then we have
y+2/y=2x+1/x since neither x,y=0
y^2x+2x=2yx^2+y

2yx^2-(y^2+2)x+y=0
x=[y^2+2+-sqrt((y^2+2)^2-8y^2)]/(4y)
x=[y^2+2+-sqrt(y^4+4y^2+4-8y^2)]/(4y)
x=[y^2+2+-sqrt(y^4-4y^2+4)]/(4y)
x=[y^2+2+-sqrt((y^2-2)^2)]/(4y)
x=[y^2+2+-(y^2-2)]/(4y)
so we have either
x=y/2 or x=1/y

now lets say we start with T0=x and we use Tk=T(k-1)/2 for each k from 1 to 1994 and then T1995=1/T1994 then we end up with
T0=x=2^1994/x and thus
x^2=2^1994
x=2^997

Thus it would appear that 2^997 is the maximum value for T0


  Posted by Daniel on 2009-09-20 15:27:47
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