To solve the recurrence relation $a_n = 2a_{n-1} + 3^{n+1}$, we can approach it by breaking it down into two parts: the homogeneous solution and a particular solution.

Step 1: Solve the Homogeneous Part

The homogeneous part of the recurrence relation is: $a_n^{(h)} = 2a_{n-1}^{(h)}$ This suggests that the solution to the homogeneous recurrence relation is of the form: $a_n^{(h)} = A \cdot 2^n$ where $A$ is a constant.

Step 2: Find a Particular Solution

To find a particular solution to the non-homogeneous recurrence $a_n = 2a_{n-1} + 3^{n+1}$, let's try a particular solution of the form: $a_n^{(p)} = B \cdot 3^n$ where $B$ is a constant to be determined.

Substitute $a_n^{(p)} = B \cdot 3^n$ into the recurrence relation:

1. The left-hand side is $a_n^{(p)} = B \cdot 3^n$.
2. The right-hand side is $2a_{n-1}^{(p)} + 3^{n+1} = 2 \cdot B \cdot 3^{n-1} + 3^{n+1}$.

Rewrite $3^{n+1}$ as $3 \cdot 3^n$: $B \cdot 3^n = 2 \cdot B \cdot 3^{n-1} + 3 \cdot 3^n$

Now divide both sides by $3^n$: $B = \frac{2B}{3} + 3$

Multiply through by 3 to clear the fraction: $3B = 2B + 9$

Solving for $B$: $B = 9$

Thus, a particular solution is: $a_n^{(p)} = 9 \cdot 3^n$

Step 3: General Solution

The general solution to the recurrence relation is the sum of the homogeneous and particular solutions: $a_n = a_n^{(h)} + a_n^{(p)} = A \cdot 2^n + 9 \cdot 3^n$

Step 4: Determine the Constant $A$

If we have an initial condition, such as $a_0$, we could use it to solve for $A$. Without an initial condition, $A$ will remain as an undetermined constant.

Final Answer

The general solution to the recurrence relation is: $a_n = A \cdot 2^n + 9 \cdot 3^n$ where $A$ is a constant determined by an initial condition.

Would you like to go over any specific part of this solution, or do you have any initial conditions for this recurrence relation?

Here are five related questions you might consider:

1. How would you find the value of $A$ if given an initial condition like $a_0 = 5$?
2. Can you verify the solution by substituting $a_n = A \cdot 2^n + 9 \cdot 3^n$ back into the recurrence?
3. How would the solution change if the recurrence relation had a different coefficient, say $a_n = 3a_{n-1} + 3^{n+1}$?
4. What are the implications of the solution behavior as $n \to \infty$ in terms of the terms $2^n$ and $3^n$?
5. How can we modify the approach if the recurrence relation includes more terms or a different type of non-homogeneous term?

Tip: Always verify your particular solution by substituting it back into the original recurrence relation to check for consistency.