T(n)=2T(n-1) +(n+5)3^n, n>=1. solve using recurrence relation method

To solve the recurrence relation \( T(n) = 2T(n-1) + (n+5)3^n \) for \( n \geq 1 \), we follow these steps:

### Step 1: Solve the Homogeneous Part
First, we focus on the homogeneous part of the recurrence relation:

\[
T(n)_{\text{homogeneous}} = 2T(n-1)
\]

Assume the solution is of the form \( T_h(n) = r^n \). Substituting this into the homogeneous recurrence relation gives:

\[
r^n = 2r^{n-1}
\]

Dividing both sides by \( r^{n-1} \):

\[
r = 2
\]

So, the homogeneous solution is:

\[
T_h(n) = A \cdot 2^n
\]

### Step 2: Find a Particular Solution
Next, we need to find a particular solution \( T_p(n) \) to the non-homogeneous equation:

\[
T(n) = 2T(n-1) + (n+5)3^n
\]

Since the non-homogeneous term is of the form \( (n+5)3^n \), we assume a particular solution of the form:

\[
T_p(n) = (an + b)3^n
\]

### Step 3: Substitute the Particular Solution
Substitute \( T_p(n) = (an + b)3^n \) into the original recurrence relation:

\[
T_p(n) = 2T_p(n-1) + (n+5)3^n
\]

Substituting:

\[
(an + b)3^n = 2 \cdot \left[ a(n-1) + b \right]3^{n-1} + (n+5)3^n
\]

Simplifying the right side:

\[
(an + b)3^n = 2a(n-1) \cdot 3^{n-1} + 2b \cdot 3^{n-1} + (n+5)3^n
\]
\[
= 2a(n-1) \cdot \frac{3^n}{3} + 2b \cdot \frac{3^n}{3} + (n+5)3^n
\]
\[
= \frac{2a(n-1)3^n}{3} + \frac{2b3^n}{3} + (n+5)3^n
\]
\[
= \frac{2a(n-1) + 2b}{3}3^n + (n+5)3^n
\]

Now, compare coefficients of like terms (the coefficients of \( n3^n \) and \( 3^n \)) on both sides:

1. Coefficient of \( n3^n \):
   \[
   a = \frac{2a}{3} + 1
   \]

   Multiply by 3 to simplify:
   \[
   3a = 2a + 3
   \]
   \[
   a = 3
   \]

2. Coefficient of \( 3^n \):
   \[
   b = \frac{-2a + 2b}{3} + 5
   \]

   Substitute \( a = 3 \):
   \[
   b = \frac{-6 + 2b}{3} + 5
   \]
   \[
   3b = -6 + 2b + 15
   \]
   \[
   b - 2b = 9
   \]
   \[
   b = -9
   \]

So the particular solution is:

\[
T_p(n) = (3n - 9)3^n
\]

### Step 4: General Solution
The general solution is the sum of the homogeneous and particular solutions:

\[
T(n) = T_h(n) + T_p(n) = A \cdot 2^n + (3n - 9)3^n
\]

### Step 5: Apply Initial Conditions (if provided)
If initial conditions are given, such as \( T(0) \) or \( T(1) \), they can be used to determine the constant \( A \). Without initial conditions, the solution remains in its general form.

### Final Answer
The solution to the recurrence relation \( T(n) = 2T(n-1) + (n+5)3^n \) is:

\[
T(n) = A \cdot 2^n + (3n - 9)3^n
\]

Would you like to explore the initial condition application, or do you have any other questions?

### Related Questions
1. How do you find the constants in the general solution using initial conditions?
2. What if the non-homogeneous term had a different form, such as \( \sin(n) \) or \( e^n \)?
3. How can you solve a recurrence relation if the characteristic equation has complex roots?
4. What is the significance of the particular solution in solving non-homogeneous recurrence relations?
5. Can you use the method of undetermined coefficients for all non-homogeneous recurrence relations?
6. How do you verify if a solution to a recurrence relation is correct?
7. How do you handle cases where the particular solution is of the same form as the homogeneous solution?
8. What happens if the recurrence relation is non-linear?

### Tip
Always double-check the particular solution by substituting it back into the original recurrence relation to ensure it satisfies the equation.

Learn how to solve the recurrence relation T(n) = 2T(n-1) + (n+5)3^n with a detailed step-by-step solution. This problem involves advanced concepts in recurrence relations, including homogeneous and particular solutions. Suitable for advanced undergraduate students.

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