To solve the recurrence relation: $a_n = 4a_{n-1} - 4a_{n-2} + (n+1)2^n$ we can split the solution into two parts:

1. Find the homogeneous solution for $a_n = 4a_{n-1} - 4a_{n-2}$.
2. Find the particular solution for $a_n = 4a_{n-1} - 4a_{n-2} + (n+1)2^n$.

Let's proceed with each part:

1. Homogeneous Solution

The homogeneous equation is: $a_n = 4a_{n-1} - 4a_{n-2}$ Assume a solution of the form $a_n = r^n$. Substitute this into the homogeneous equation:

$r^n = 4r^{n-1} - 4r^{n-2}$ Dividing by $r^{n-2}$ (assuming $r \neq 0$): $r^2 = 4r - 4$ Rearrange to get the characteristic polynomial: $r^2 - 4r + 4 = 0$ This can be factored as: $(r - 2)^2 = 0$ So, we have a repeated root $r = 2$. This implies that the general form of the homogeneous solution is: $a_n^{(h)} = (c_1 + c_2 n) 2^n$ where $c_1$ and $c_2$ are constants.

2. Particular Solution

For the particular solution, we look at the non-homogeneous term, which is $(n+1)2^n$.

Since $2^n$ is already a solution to the homogeneous equation (due to the characteristic root $r = 2$), we will try a particular solution of the form: $a_n^{(p)} = (An^2 + Bn + C) 2^n$ Substitute $a_n^{(p)}$ into the original recurrence relation to determine $A$, $B$, and $C$.

1. Calculate $a_{n-1}^{(p)} = (A(n-1)^2 + B(n-1) + C)2^{n-1}$.
2. Calculate $a_{n-2}^{(p)} = (A(n-2)^2 + B(n-2) + C)2^{n-2}$.

By plugging these expressions into the recurrence and equating coefficients of $n^2$, $n$, and constants, we can solve for $A$, $B$, and $C$.

After finding these values, the general solution to the recurrence is: $a_n = a_n^{(h)} + a_n^{(p)} = (c_1 + c_2 n)2^n + (An^2 + Bn + C)2^n$ where $c_1$ and $c_2$ are determined by initial conditions.