Solve the recurrence relation an=4an-1 - 4an-2 +(n+1)2^n

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.

Solve the recurrence relation an = 4an-1 - 4an-2 + (n+1)2^n

This solution walks through solving the recurrence relation an = 4an-1 - 4an-2 + (n+1)2^n. Learn to find the homogeneous and particular solutions, apply the characteristic polynomial, and combine solutions to get the general answer. Suitable for college-level students studying recurrence relations.

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