- #1
Andrax
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Homework Statement
we have u(1)=1
u(n+1)=[itex]\frac{1}{16}[/itex](1+4u(n)+[itex]\sqrt{1+24u(n)}[/itex])
v(n)^2=1+24u(n)
Prove that [itex]\foralln[/itex][itex]\inN*[/itex]: v(n+1)-3=[itex]\frac{1}{2}[/itex](v(n)-3)
Homework Equations
The Attempt at a Solution
by working on v(n+1)-3 and replacing it by [itex]\sqrt{1+24u(n+1)}[/itex] didn't get me anywhere just a bunch of calc
also i made 6u(n+1)=[itex]\frac{1}{16}[/itex](v(n)+[itex]\frac{1}{2}[/itex])^2+[itex]\frac{19}{64}[/itex] but it didn't work for me , also just to mention the progression is not arithmetic or anything it's just (u(n)):n[itex]\geq1[/itex]
PS: the "^2" means squared
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