Prove Progression Proof of v(n+1)-3

[Andrax]

Homework Statement

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)

Homework Equations

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

 

[haruspex]

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

 

[Andrax]

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

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

 

[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
v_n-3=(v_n-1-3)/2
i.e.
v_n=(v_n-1+3)/2
Now work with v_n+1² and see if you can show it equal to ((v_n+3)/2)². (Work from both ends.)

 

[Andrax]

Thanks solved it.