Understanding the Recurrence Relation

The given problem asks for the general solution of a linear, non-homogeneous recurrence relation:

\(a_r - 5a_{r-1} + 6a_{r-2} = 4^r\), for \(r \ge 2\).

A recurrence relation defines a sequence where each term is a function of the preceding terms. To find the general solution of a non-homogeneous linear recurrence relation, we need to find two parts:

1. The homogeneous solution, denoted as \(a_r^{(h)}\).
2. The particular solution, denoted as \(a_r^{(p)}\).

The general solution is the sum of these two parts: \(a_r = a_r^{(h)} + a_r^{(p)}\).

Finding the Homogeneous Solution

The homogeneous part of the recurrence relation is obtained by setting the non-homogeneous term (\(4^r\)) to zero:

\(a_r - 5a_{r-1} + 6a_{r-2} = 0\).

To solve this homogeneous recurrence relation, we write its characteristic equation. We assume a solution of the form \(a_r = k^r\). Substituting this into the homogeneous equation gives:

\(k^r - 5k^{r-1} + 6k^{r-2} = 0\)

Assuming \(k \ne 0\) and dividing by \(k^{r-2}\) (for \(r \ge 2\)), we get the characteristic equation:

\(k^2 - 5k + 6 = 0\)

Now, we solve this quadratic equation for \(k\). We can factor it:

\((k - 2)(k - 3) = 0\)

The roots are \(k_1 = 2\) and \(k_2 = 3\). Since the roots are distinct real numbers, the homogeneous solution is of the form:

\(a_r^{(h)} = A_1 \cdot k_1^r + A_2 \cdot k_2^r\)

Substituting the roots, we get the homogeneous solution:

\(a_r^{(h)} = A_1 \cdot 2^r + A_2 \cdot 3^r\)

where \(A_1\) and \(A_2\) are arbitrary constants determined by initial conditions (which are not provided in this problem asking for the general solution).

Finding the Particular Solution

The non-homogeneous term is \(f(r) = 4^r\). Since the base of this exponential (4) is not a root of the characteristic equation (the roots are 2 and 3), we assume a particular solution of the form:

\(a_r^{(p)} = C \cdot 4^r\)

where \(C\) is a constant that we need to find. We substitute this assumed form into the original non-homogeneous recurrence relation:

\(a_r - 5a_{r-1} + 6a_{r-2} = 4^r\)

\((C \cdot 4^r) - 5(C \cdot 4^{r-1}) + 6(C \cdot 4^{r-2}) = 4^r\)

To solve for \(C\), we can factor out \(C \cdot 4^{r-2}\) or divide the entire equation by \(4^{r-2}\) (for \(r \ge 2\)):

\(C \cdot \dfrac{4^r}{4^{r-2}} - 5C \cdot \dfrac{4^{r-1}}{4^{r-2}} + 6C \cdot \dfrac{4^{r-2}}{4^{r-2}} = \dfrac{4^r}{4^{r-2}}\)

\(C \cdot 4^2 - 5C \cdot 4^1 + 6C \cdot 4^0 = 4^2\)

\(16C - 20C + 6C = 16\)

\((16 - 20 + 6)C = 16\)

\(2C = 16\)

\(C = \dfrac{16}{2}\)

\(C = 8\)

So, the particular solution is:

\(a_r^{(p)} = 8 \cdot 4^r\)

Forming the General Solution

The general solution is the sum of the homogeneous and particular solutions:

\(a_r = a_r^{(h)} + a_r^{(p)}\)

\(a_r = (A_1 \cdot 2^r + A_2 \cdot 3^r) + (8 \cdot 4^r)\)

The general solution of the recurrence relation \(a_r - 5a_{r-1} + 6a_{r-2} = 4^r\) is:

\(a_r = A_1 \cdot 2^r + A_2 \cdot 3^r + 8 \cdot 4^r\)

where \(A_1\) and \(A_2\) are arbitrary constants.

Comparing with Options

Let's compare our derived general solution \(A_1 \cdot 2^r + A_2 \cdot 3^r + 8 \cdot 4^r\) with the given options:

• Option 1: \(A_1.2^r+A_2.3^r-\dfrac{1}{8}.4^r\) - Incorrect particular solution term.
• Option 2: \(A_1.5^r+A_2.6^r+8.4^r\) - Incorrect homogeneous solution terms (roots 5 and 6 instead of 2 and 3).
• Option 3: \(A_1.2^r+A_2.3^r+8.4^r\) - Matches our derived general solution.
• Option 4: \(A_1.2^r+A_2.4^r+\dfrac{1}{8}.3^r\) - Incorrect homogeneous and particular solution terms.

Therefore, the correct option is the one that matches the derived general solution \(A_1.2^r+A_2.3^r+8.4^r\).

Revision Table: Recurrence Relation Solution Steps

Additional Information: Solving Recurrence Relations

Linear constant-coefficient recurrence relations are similar to linear constant-coefficient differential equations. The methods for finding homogeneous and particular solutions have parallels.

Case for Particular Solution Guess:

The form of the particular solution \(a_r^{(p)}\) depends on the form of the non-homogeneous term \(f(r)\) and the roots of the characteristic equation.

• If \(f(r) = P(r)\), a polynomial in \(r\) of degree \(d\), and 1 is not a root of the characteristic equation, guess \(a_r^{(p)}\) as a general polynomial of degree \(d\).
• If \(f(r) = c^r \cdot P(r)\), where \(P(r)\) is a polynomial of degree \(d\):
  • If \(c\) is not a root of the characteristic equation, guess \(a_r^{(p)} = c^r \cdot Q(r)\), where \(Q(r)\) is a general polynomial of degree \(d\).
  • If \(c\) is a root of the characteristic equation with multiplicity \(m\), guess \(a_r^{(p)} = r^m \cdot c^r \cdot Q(r)\), where \(Q(r)\) is a general polynomial of degree \(d\).

In our case, \(f(r) = 4^r = 4^r \cdot 1\). Here, \(c=4\) and \(P(r)=1\) (a polynomial of degree \(d=0\)). Since 4 is not a root of the characteristic equation, we guessed \(a_r^{(p)} = 4^r \cdot C\), where \(C\) is a polynomial of degree 0 (a constant).

Importance of General Solution:

The general solution includes arbitrary constants (\(A_1, A_2\)). These constants can be uniquely determined if initial conditions (e.g., values of \(a_0, a_1\)) are provided. Without initial conditions, the solution represents the entire family of sequences that satisfy the recurrence relation.