91Ó°ÊÓ

To begin solving the recurrence relation, we first address its homogeneous part. The homogeneous part helps us understand the part of the solution that repeats uniformly without any additional external input.

Given the homogeneous part of the recurrence relation:
\(a_n = 2a_{n-1}\)
we assume a solution of the form \(a_n = r^n\). To determine \(r\), we substitute \(r^n\) into the equation:
\(r^n = 2r^{n-1}\)
Dividing both sides by \( r^{n-1} \) gives:
\(r = 2\)
This implies that the solution to the homogeneous part is of the form:
\(a_n^{(h)} = C \times 2^n\),
where \(C\) is a constant. This homogeneous solution captures the repetitive, multiplying nature of the sequence.

Next, we need to find the particular solution of the recurrence relation. This solution accounts for the non-homogeneous term, which in this case is \(2n^2\). The particular solution is found by guessing a form that can accommodate this non-homogeneous term.

We assume a particular solution of the form:
\(a_n^{(p)} = An^2 + Bn + D\)
We substitute this into the original recurrence relation and balance the equation:
\(An^2 + Bn + D = 2(A(n-1)^2 + B(n-1) + D) + 2n^2\)
By simplifying and collecting like terms, we derive equations for \(A\), \(B\), and \(D\).
Through equating the coefficients of similar powers of \(n\), we obtain:
\(A = -2\)
\(B = 8\)
\(D = -14\)
Thus, the particular solution takes the form:
\(a_n^{(p)} = -2n^2 + 8n - 14\).

To find a specific solution to the recurrence relation, we need to apply an initial condition, given here as \(a_1 = 4\).

We use the general solution we derived, combining both the homogeneous and particular solutions:
\(a_n = C \times 2^n - 2n^2 + 8n - 14\).
To find the constant \(C\), we substitute \(n = 1\) as follows:
\(4 = C \times 2^1 - 2(1^2) + 8(1) - 14\)
Simplifying this equation:
\(4 = 2C - 2 + 8 - 14\)
\(4 = 2C - 8\)
\(12 = 2C\)
\(C = 6\).
Therefore, our initial condition helps us tailor the general solution to a specific instance.

A non-homogeneous recurrence relation includes a term beyond the repetitive components seen in homogeneous relations. For our problem, this term is \(2n^2\).

The general form of these relations is:
\(a_n = fa_{n-1} + g(n)\),
where \(g(n)\) represents the non-homogeneous part. Finding solutions involves two steps:

• Solving the homogeneous part, which assumes \(g(n) = 0\).
• Finding a particular solution that accommodates \(g(n)\).

The final step combines these two solutions into a comprehensive general solution.

The general solution to a non-homogeneous recurrence relation is the sum of the homogeneous and particular solutions. This combination captures all potential behaviors of the sequence.

From our homogeneous part, we have:
\(a_n^{(h)} = C \times 2^n\)
For the particular part, we derived:
\(a_n^{(p)} = -2n^2 + 8n - 14\)
The general solution merging these parts is:
\(a_n = C \times 2^n - 2n^2 + 8n - 14\).
Using the initial condition \(a_1 = 4\), we found the specific value of \(C = 6\).

Finally, substituting \(C\) back, we get:
\(a_n = 6 \times 2^n - 2n^2 + 8n - 14\). This is our complete solution for the given recurrence relation and initial condition.