Homework Help: Progression proof

1. Dec 24, 2012

Andrax

1. The problem statement, all variables and given/known data
we have u(1)=1
u(n+1)=$\frac{1}{16}$(1+4u(n)+$\sqrt{1+24u(n)}$)

v(n)^2=1+24u(n)

Prove that $\foralln$$\inN*$: v(n+1)-3=$\frac{1}{2}$(v(n)-3)
2. Relevant equations

3. The attempt at a solution
by working on v(n+1)-3 and replacing it by $\sqrt{1+24u(n+1)}$ didn't get me anywhere just a bunch of calc
also i made 6u(n+1)=$\frac{1}{16}$(v(n)+$\frac{1}{2}$)^2+$\frac{19}{64}$ but it didn't work for me , also just to mention the progression is not arithmetic or anything it's just (u(n)):n$\geq1$
PS: the "^2" means squared

Last edited: Dec 24, 2012
2. Dec 24, 2012

haruspex

Have you tried induction? Btw, there's a parenthesis missing in un+1=(1/16)(1+4un+√(1+24un). Should it be un+1=(1/16)(1+4un)+√(1+24un)?

3. Dec 24, 2012

Andrax

sorry forgot the parenthesis , using induction will make things worse ethe problem is the square root it's causing the problem here..

4. Dec 24, 2012

haruspex

Induction doesn't look too bad. Don't get tangled up with square roots - work with the squared-up form. Assume true up to
vn-3=(vn-1-3)/2
i.e.
vn=(vn-1+3)/2
Now work with vn+12 and see if you can show it equal to ((vn+3)/2)2. (Work from both ends.)

5. Dec 25, 2012

Andrax

Thanks solved it.