# 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.