Question
Then is equal to
Options
Solution
Key Concepts and Formulas
- Linear Non-Homogeneous Recurrence Relations: A recurrence relation of the form . The general solution is the sum of the homogeneous solution and a particular solution.
- Homogeneous Solution: For , the characteristic equation is . The form of the homogeneous solution depends on the roots of this equation.
- Particular Solution: For a constant non-homogeneous term , if is not a root of the characteristic equation, a particular solution is of the form . If is a simple root, . If is a double root, .
- Algebraic Factoring: Simplifying complex expressions by grouping terms and finding common factors.
Step-by-Step Solution
Step 1: Analyze the Expression to be Evaluated The expression we need to evaluate is . We observe that this expression can be factored by grouping. This factorization simplifies the problem significantly. Our goal is now to find the values of for specific values of .
Step 2: Find the Closed-Form Expression for the Sequence The given recurrence relation is . First, consider the homogeneous part: . The characteristic equation is . Factoring the quadratic, we get . The roots are and . The homogeneous solution is .
Next, find a particular solution for the non-homogeneous term '+1'. Since 1 is a root of the characteristic equation, we guess a particular solution of the form . Substitute this into the recurrence relation: . So, the particular solution is .
The general solution is the sum of the homogeneous and particular solutions: .
Step 3: Use Initial Conditions to Determine Constants A and B We are given and . For : . For : . Substitute into the second equation: . Since , we have . Thus, the closed-form expression for the sequence is .
Step 4: Evaluate the Factored Expression We need to evaluate . Let's find a general form for . Using the closed form : So, .
Now, we can substitute the values of into this result: For the first factor, , we have . So, . For the second factor, , we have . So, .
The expression is the product of these two values:
Common Mistakes & Tips
- Incorrectly solving the characteristic equation or finding the particular solution: Double-check the roots and the form of the particular solution based on whether the non-homogeneous term's root is a root of the characteristic equation.
- Algebraic errors in simplification: Carefully perform the substitutions and algebraic manipulations, especially when dealing with exponents.
- Forgetting to use initial conditions: The constants A and B are crucial for obtaining the correct closed-form expression, which is necessary for the final calculation.
- Not factoring the target expression: Trying to calculate individual terms and then substituting them into the original expression would be extremely tedious and error-prone.
Summary The problem involves a linear non-homogeneous recurrence relation. We first factored the expression to be evaluated into . Then, we found the closed-form expression for the sequence by solving the homogeneous and particular parts of the recurrence relation and using the initial conditions. We derived a simplified form for as . Finally, we substituted the appropriate values of to calculate the two factors and their product, yielding the final answer.
The final answer is .