Solve the Recurrence Relation a_n = 4a_{n-1} + 16^{n-1}

In this post, we will learn how to solve the linear non-homogeneous recurrence relation \( a_n = 4a_{n-1} + 16^{n-1} \) with the initial condition \(a_1 = 1 \).

Solution:

This is a linear non-homogeneous recurrence relation with constant coefficients. To solve it, we will find the general solution to its associated linear homogeneous equation and then find a particular solution to the given non-homogeneous equation.

  • Step 1: Solve the Homogeneous Equation

The associated linear homogeneous recurrence relation is \( a_n = 4a_{n-1} \). The characteristic equation of this recurrence relation is \( r^n = 4r^{n-1} \implies r=4\).

The characteristic root \( r = 4 \) gives the general solution of this recurrence relation :

\( a_n^{(h)} = A \cdot 4^n \) where \( A \) is a constant.

  • Step 2: Find a Particular Solution

We now find a particular solution. For the non-homogeneous term \( 16^{n-1} \), guess a solution of the form: \( a_n^{(p)} = B \cdot 16^n \) where \( B \) is a constant.

Substituting into the recurrence, we get the following:

\( B \cdot 16^n = 4 \cdot B \cdot 16^{n-1} + 16^{n-1} \)

Now we need to simplify it for the solution:

\( B \cdot 16^n = \frac{4B}{16} \cdot 16^n + \frac{1}{16} \cdot 16^n \\ \Rightarrow B = \frac{B}{4} + \frac{1}{16} \\ \Rightarrow B\, – \frac{B}{4} = \frac{1}{16} \\ \Rightarrow \frac{3B}{4} = \frac{1}{16} \\ \Rightarrow B = \frac{1}{12} \\ \)

Thus, the particular solution is: \( a_n^{(p)} = \frac{16^n}{12} \)

  • Step 3: General Solution

Combine the homogeneous and particular solutions:

\( a_n = A \cdot 4^n + \frac{16^n}{12} \)

Using \( a_1 = 1 \):

\( 1 = A \cdot 4^1 + \frac{16^1}{12} \\ \Rightarrow 1 = 4A + \frac{4}{3} \\ \Rightarrow 4A = 1 – \frac{4}{3} = -\frac{1}{3} \\ \Rightarrow A = -\frac{1}{12} \\ \)

Substitute \( A \) back into the general solution:

\( \displaystyle a_n = -\frac{1}{12} \cdot 4^n + \frac{16^n}{12} = \frac{16^n\, – 4^n}{12} \)

Therefore, the solution to the recurrence relation is:

\(\displaystyle \boxed{a_n = \frac{16^n\, – 4^n}{12}} \)

Please let me know in the comments if you find any error in this solution.

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