Solve the recurrence relation a n =2a n - 1 +2^ n alpha_{0} = 2


To solve the recurrence relation \( a_n = 2a_{n-1} + 2^n \) with the initial condition \( a_0 = 2 \), follow these steps:

### Step 1: Homogeneous Solution
The first step is to solve the associated homogeneous recurrence relation:
\[
a_n = 2a_{n-1}
\]
This is a simple first-order recurrence relation. Assuming a solution of the form \( a_n = r^n \), we substitute into the homogeneous equation:
\[
r^n = 2r^{n-1}
\]
This simplifies to:
\[
r = 2
\]
Thus, the solution to the homogeneous equation is:
\[
a_n^{(h)} = C \cdot 2^n
\]
where \( C \) is a constant to be determined later.

### Step 2: Particular Solution
Next, we find a particular solution to the non-homogeneous recurrence relation:
\[
a_n = 2a_{n-1} + 2^n
\]
Since the non-homogeneous term is \( 2^n \), we try a particular solution of the form \( a_n^{(p)} = A \cdot n \cdot 2^n \), where \( A \) is a constant to be determined. Substituting this into the recurrence relation:

\[
A \cdot n \cdot 2^n = 2(A \cdot (n-1) \cdot 2^{n-1}) + 2^n
\]

Simplify the terms on the right-hand side:

\[
A \cdot n \cdot 2^n = 2A \cdot (n-1) \cdot 2^{n-1} + 2^n
\]
\[
A \cdot n \cdot 2^n = A \cdot (n-1) \cdot 2^n + 2^n
\]
\[
A \cdot n = A \cdot (n-1) + 1
\]
\[
A \cdot n = A \cdot n - A + 1
\]
This simplifies to:
\[
0 = -A + 1
\]
Thus, \( A = 1 \).

So the particular solution is:
\[
a_n^{(p)} = n \cdot 2^n
\]

### Step 3: General Solution
The general solution to the recurrence relation is the sum of the homogeneous and particular solutions:
\[
a_n = C \cdot 2^n + n \cdot 2^n
\]
\[
a_n = (C + n) \cdot 2^n
\]

### Step 4: Apply Initial Condition
To find \( C \), use the initial condition \( a_0 = 2 \):
\[
a_0 = (C + 0) \cdot 2^0 = C = 2
\]

Thus, \( C = 2 \), and the final solution is:
\[
a_n = (2 + n) \cdot 2^n
\]

### Final Answer:
\[
a_n = (2 + n) \cdot 2^n
\]

Would you like further details or have any questions? 

Here are 5 related questions for practice:
1. Solve the recurrence relation \( a_n = 3a_{n-1} + 2^n \), with \( a_0 = 1 \).
2. Find the general solution for \( a_n = 4a_{n-1} + 3^n \), with \( a_0 = 0 \).
3. Solve the recurrence relation \( a_n = a_{n-1} + n^2 \), with \( a_0 = 5 \).
4. Determine the solution for \( a_n = 5a_{n-1} + n \), with \( a_0 = 10 \).
5. Solve the recurrence relation \( a_n = 2a_{n-1} + n \), with \( a_0 = 3 \).

**Tip:** For non-homogeneous recurrence relations, always try a particular solution form that mirrors the non-homogeneous term (e.g., polynomials for linear terms, exponentials for exponential terms).

Solve the recurrence relation a_n = 2a_{n-1} + 2^n, a_0 = 2

This page provides a step-by-step solution to the recurrence relation a_n = 2a_{n-1} + 2^n with the initial condition a_0 = 2. It covers key concepts such as solving homogeneous and non-homogeneous recurrence relations and applying initial conditions. Suitable for undergraduate students studying discrete mathematics.

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