<s>[INST] How to find the general term of the following sequence? Consider the following recurrence problem:
\begin{align}
d_{i-1} &= 2\varphi_{i+1}+4\varphi_i + 8d_i-7d_{i+1} - F \left( \delta_{i,N} + \delta_{i,N+1} \right) \, , \\
\varphi_{i-1} &= -7\varphi_{i+1}-16\varphi_{i} + 24 \left( d_{i+1}-d_{i} \right) + F \left( \delta_{i,N} + \delta_{i,N+1} \right) \, , 
\end{align}
where $d_i, i \in \{2, \cdots , N+1\}$ represent displacements, $\varphi_i$ inclinations, and $F$ is a known force acting at the nodes $N$ and $N+1$.
We require by the system symmetry that $d_{N+1}=d_N = d_\mathrm{C}$ and $\varphi_N = -\varphi_{N+1}= \varphi_\mathrm{C}$, where $d_\mathrm{C}$ and $\varphi_\mathrm{C}$ are still to be determined from the boundary conditions:
$d_1 = 0$ (zero displacement) and $2\varphi_1+\varphi_2 = 3d_2$ (zero torque)
In order to proceed, i have tried to first determine $d_{N-1}$ and $\varphi_{N-1}$ from the system above, and then $d_{N-2}$ and $\varphi_{N-2}$, etc... in a recursive way and then try to find out the general term of the resulting sequences. 
For the term $N-1$, we obtain
\begin{align}
d_{N-1} &= d_\mathrm{C}+2\varphi_\mathrm{C}-F \, \\
\varphi_{N-1} &= -9\varphi_\mathrm{C}+3F \, .
\end{align}
Analogously, we get for the term $N-2$
\begin{align}
d_{N-2} &= d_\mathrm{C}-18\varphi_\mathrm{C}+4F \, \\
\varphi_{N-2} &= 89\varphi_\mathrm{C}-24F \, .
\end{align}
I was wondering whether there is a particular way to figure out the general term of such sequence.
Any help or suggestions are most welcome.
 [/INST] Thanks again to Yuri for the helpful insights. I can say that I have now found the solution to my problem.
First of all, let us define $D_i = d_i-d_{i-1}$, in the same way as Yuri did, and let us get a recurrence equation that involves $\varphi_i$ only. 
Actually, original problem (which is equivalent to the one stated above) was
\begin{align}
d_{i+1}-2d_i+d_{i-1} - \frac{1}{4} \left( \varphi_{i+1}-\varphi_{i-1}\right) + F \left( \delta_{i, N} + \delta_{i, N+1} \right) &=0\, , \\
d_{i+1}-d_{i-1} - \frac{1}{3} \left( \varphi_{i+1}+4\varphi_i+\varphi_{i-1} \right) &= 0 \, .
\end{align}
Accordingly, it can easily be shown that for $0 <i <N$,
\begin{align}
D_{i+1}-D_i &= \frac{1}{4} \left( \varphi_{i+1}-\varphi_{i-1} \right) \, , \\
D_{i+1}+D_i &= \frac{1}{3} \left( \varphi_{i+1}+4\varphi_i+\varphi_{i-1} \right) \, .
\end{align}
This leads to
\begin{align}
D_i = \frac{7}{24} \varphi_i + \frac{1}{24} \varphi_{i-2} + \frac{2}{3} \varphi_{i-1}  =
\frac{1}{24} \varphi_{i+1}+\frac{7}{24} \varphi_{i-1} + \frac{2}{3} \varphi_i  \, , \tag{1} \label{1}
\end{align}
or 
$$
\frac{1}{24} \left( \varphi_{i+1}-\varphi_{i-2} \right)
+\frac{3}{8} \left( \varphi_i-\varphi_{i-1} \right) =0\, .
$$
Using Yuri's powerful approach, and posing $\varphi_i = A/p^i$ yields
$$
p^3+9p^2-9p-1=0 \, , 
$$
whose solutions are $p=1$ and $p=-5\pm 2 \sqrt{6}$.
Finally, 
$$
\varphi_i = C_1 + C_2 \left(-5-2\sqrt{6}\right)^i + C_3 \left(-5+2\sqrt{6}\right)^i \, .
$$
(again, by noting that $\left(-5-2\sqrt{6}\right)\left(-5+2\sqrt{6}\right)=1$.)
Therefore, 
$$
d_i 
=D_i + D_{i-1} + \cdots + D_2 =
\sum_{n=2}^{n=i} D_n \, , 
$$
where the expression of $D_n$ follows readily from Eq. \eqref{1}, and using the fact that $d_1=0.$
The final expression read
$$
D_i = C_1 + 
C_2 \left( -1+\frac{\sqrt{6}}{2} \right) \left( -5-2\sqrt{6} \right)^i + 
C_3 \left( -1-\frac{\sqrt{6}}{2} \right) \left( -5+2\sqrt{6} \right)^i \, ,
$$
and
$$
d_i = (i-1)C_1 + 
C_2 \left( -1+\frac{5\sqrt{6}}{12} \right)
\left[ 49+20\sqrt{6}-\left( -5-2\sqrt{6} \right)^{i+1} \right] + 
C_3 \left( 1+\frac{5\sqrt{6}}{12} \right)
\left[ -49+20\sqrt{6}+\left( -5+2\sqrt{6} \right)^{i+1} \right] \, .
$$
Clearly, $d_1=0$.
Finally, at this point, the 3 unknown coefficients can readily be determined from the 3 boundary conditions.

I have checked that the solution is very well behaved and is in full agreement with the numerical solution. 

</s>