Prove by induction $u_{n+2}+u_n=4u_{n+1}$ for all $n\in \mathbb{N}$, where $u_n=(2+\sqrt{3})^n+(2-\sqrt{3})^n$. For $n=1$,
My efforts:
LHS= $u_3+u_1=(2+\sqrt{3})^3+(2-\sqrt{3})^3+ (2+\sqrt{3})^1+(2-\sqrt{3})^1=56$.
RHS= $4u_2=4\left[(2+\sqrt{3})^2+(2-\sqrt{3})^2 \right]=56$.
Let us assume that the result is true for $n=k$ for some $k\in \mathbb{N}$.
Then $u_{k+2}+u_k=4u_{k+1}$.
Now, $u_{k+3}+u_{k+1}=4u_{k+1}$
How can I show the result for $n=k+1$?
To make it a bit easier to deal with, the relation you're asked to prove is true for all $n \in \mathbb{N}$ can be rewritten as
$$u_{n+2} = 4u_{n+1} - u_{n} \tag{1}\label{eq1A}$$
You're given that
$$u_n = (2 + \sqrt{3})^n + (2 - \sqrt{3})^n \tag{2}\label{eq2A}$$
for all $n \in \mathbb{N}$. Using \eqref{eq2A} to define $u_1$ and $u_2$, you've done the base case by proving that \eqref{eq1A} is true for $u_3$. Next, use strong induction to assume \eqref{eq1A} is true for all $n \le k$ for some $k \ge 1$. Now, you need to use the values in \eqref{eq2A} of $u_j$ where $j \le k + 2$ to show \eqref{eq1A} is also true for $n = k + 1$. To do this, since the RHS of \eqref{eq1A} can be expressed using those terms, simplify its value as shown below to get
$$\begin{equation}\begin{aligned} 4u_{k+2} - u_{k+1} & = 4\left((2 + \sqrt{3})^{k+2} + (2 - \sqrt{3})^{k+2}\right) - \left((2 + \sqrt{3})^{k+1} + (2 - \sqrt{3})^{k+1}\right) \\ & = 4(2 + \sqrt{3})^{k+2} - (2 + \sqrt{3})^{k+1} + 4(2 - \sqrt{3})^{k+2} - (2 - \sqrt{3})^{k+1} \\ & = (4(2 + \sqrt{3}) - 1)(2 + \sqrt{3})^{k+1} + (4(2 - \sqrt{3}) - 1)(2 - \sqrt{3})^{k+1} \\ & = (7 + 4\sqrt{3})(2 + \sqrt{3})^{k+1} + (7 - 4\sqrt{3})(2 - \sqrt{3})^{k+1} \\ & = (2 + \sqrt{3})^2(2 + \sqrt{3})^{k+1} + (2 - \sqrt{3})^2(2 - \sqrt{3})^{k+1} \\ & = (2 + \sqrt{3})^{k+3} + (2 - \sqrt{3})^{k+3} \\ & = u_{k+3} \end{aligned}\end{equation}\tag{3}\label{eq3A}$$
This shows the RHS is equal to the LHS of \eqref{eq1A}, so it's also true for $n = k + 1$. Thus, by induction, \eqref{eq1A} is true for all $n \in \mathbb{N}$.
