Question

What is the solution of the recurrence relation \[ a_n=6a_{n-1}-9a_{n-2} \] with initial conditions \[ a_0=1,\qquad a_1=6? \]

• A. \(a_n=3^n+n\times 3^n\)
• B. \(a_n=1-2^n+2\times 3^n\)
• C. \(a_n=2^n+3^n\)
• D. \(a_n=3^n\)

Hint:

For a repeated root \(r\) of multiplicity 2, the solution is \[ a_n=(C_1+C_2n)r^n. \] Always remember the extra factor \(n\) for repeated roots.

Answer 1

The Correct Option is A

Concept: A linear homogeneous recurrence relation with constant coefficients can be solved using its characteristic equation. For a recurrence relation of the form \[ a_n=c_1a_{n-1}+c_2a_{n-2}, \] we assume a solution of the form \[ a_n=r^n. \] Substituting this trial solution gives the characteristic equation. The roots of this equation determine the general form of the solution. If a root is repeated, say \(r\), then the corresponding solution is \[ a_n=(C_1+C_2n)r^n. \]

Step 1: Form the characteristic equation.
Given \[ a_n=6a_{n-1}-9a_{n-2}. \] Assume \[ a_n=r^n. \] Substituting, \[ r^n=6r^{n-1}-9r^{n-2}. \] Dividing by \(r^{n-2}\), \[ r^2-6r+9=0. \]

Step 2: Find the roots.
Factorizing, \[ r^2-6r+9=(r-3)^2. \] Thus the characteristic equation has a repeated root \[ r=3. \]

Step 3: Write the general solution.
Since the root is repeated, \[ a_n=(C_1+C_2n)3^n. \]

Step 4: Use the initial condition \(a_0=1\).
Putting \(n=0\), \[ 1=(C_1+0)3^0. \] Hence, \[ C_1=1. \]

Step 5: Use the initial condition \(a_1=6\).
Putting \(n=1\), \[ 6=(1+C_2)3. \] \[ 2=1+C_2. \] \[ C_2=1. \]

Step 6: Substitute the constants.
Therefore, \[ a_n=(1+n)3^n. \] or \[ a_n=3^n+n3^n. \] \[ \boxed{a_n=3^n+n3^n} \] Hence option (A) is correct.