Proving a Recursive Sequence Identity

Question

I am trying to prove the following identity for a recursive sequence:

$$V_{n+1} - 3 = \frac{1}{2} \cdot (V_n - 3)$$

where $U$ is a sequence defined as follows:

$$U_1 = 1$$

$$U_{n+1} = \frac{1}{16} \cdot (1+4U_n + \sqrt{1+24U_n})$$

and $V_n$ is defined as:

$$V_n = \sqrt{1+24U_n}$$

So I tried to make the exp like this:

$$U_{n+1} = \frac{1}{16} \cdot (1+4U_n + V_n)$$

I have been working on this problem for a while, but I am having trouble making any more progress. I need a hint.

Thank you in advance for your help!

Answer 1

Start with \begin{eqnarray*} 16U_{n+1} =1+ 4U_n +\sqrt{1+24 U_n}. \end{eqnarray*} Multiply by $6$ and add $4$ to both sides \begin{eqnarray*} 4(1+ 24U_{n+1}) =10+ 24U_n +6\sqrt{1+24 U_n} \end{eqnarray*} Now $10=1+9$ and the RHS is a perfect square of $(\sqrt{1+24 U_n}+3)$.

Answer 2

One easily checks by induction that $$1+24U_n=\left(3+\frac1{2^{n-2}}\right)^2.$$ From this, we deduce $$V_n=3+\frac1{2^{n-2}},$$ whence the recurrence relation on $V_n.$