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Solve the recurrence relation \(a_{n}=-5 a_{n-1}+6 a_{n-2}\), \(n \geq 2\), given \(a_{0}=5, a_{1}=19\)

Short Answer

Expert verified
The solution is \(a_n = 7 - 2(-6)^n\).

Step by step solution

01

Identify the Characteristic Equation

For the recurrence relation \(a_n = -5a_{n-1} + 6a_{n-2}\), identify the characteristic equation from the homogeneous part. Replace \(a_n\) with \(r^n\) to derive the equation: \(r^n = -5r^{n-1} + 6r^{n-2}\). Divide throughout by \(r^{n-2}\), assuming \(r eq 0\), to obtain the characteristic equation: \(r^2 + 5r - 6 = 0\).
02

Solve the Characteristic Equation

Solve the quadratic characteristic equation \(r^2 + 5r - 6 = (r - 1)(r + 6) = 0\). The roots of the equation are \(r = 1\) and \(r = -6\). This gives us the general solution for the recurrence as \(a_n = A(1)^n + B(-6)^n\).
03

Apply Initial Conditions

Use the initial conditions \(a_0 = 5\) and \(a_1 = 19\) to find the coefficients \(A\) and \(B\). For \(n = 0\), \(a_0 = A(1)^0 + B(-6)^0 = A + B = 5\). For \(n = 1\), \(a_1 = A(1)^1 + B(-6)^1 = A - 6B = 19\). This gives us the system of equations: \[A + B = 5\] \[A - 6B = 19\].
04

Solve the System of Equations

Solve the system \(A + B = 5\) and \(A - 6B = 19\). Subtract the first equation from the second: \((A - 6B) - (A + B) = 19 - 5\), which simplifies to \(-7B = 14\), giving \(B = -2\). Substituting \(B = -2\) in the first equation, \(A - 2 = 5\), and solving for \(A\) gives \(A = 7\).
05

Write the Particular Solution

Substitute \(A = 7\) and \(B = -2\) back into the general solution \(a_n = A(1)^n + B(-6)^n\), we get the particular solution: \(a_n = 7(1)^n - 2(-6)^n = 7 - 2(-6)^n\). Hence, the solution to the recurrence relation is \(a_n = 7 - 2(-6)^n\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Characteristic Equation
When dealing with linear recurrence relations like the given problem, the characteristic equation is a key concept. It provides a way to identify a pattern or formula that fits all terms in the sequence. To derive the characteristic equation, we start by assuming solutions of the form \( a_n = r^n \), where \( r \) is a constant. This approach works well for linear homogeneous recurrence relations with constant coefficients.

By substituting \( r^n \) into the given relation \( a_n = -5a_{n-1} + 6a_{n-2} \), and then dividing through by \( r^{n-2} \), we obtain a quadratic equation: \( r^2 + 5r - 6 = 0 \). This equation, known as the characteristic equation, serves as a foundational tool to determine the pattern of the sequence.
Initial Conditions
Initial conditions are essential for finding the particular solution of a recurrence relation. They provide specific values for the first few terms of the sequence, ensuring the solution meets certain criteria. In our example, the initial conditions are \( a_0 = 5 \) and \( a_1 = 19 \).

These initial conditions allow us to calculate the constants in the general solution, tailored specifically to the problem at hand. Without initial conditions, we can only find a general form of the sequence. These starting values create a unique path for how the sequence will evolve. Once we have the initial conditions, they are plugged into the general solution, forming a "system of equations" which we'll solve next.
General Solution
The general solution is a formula or expression that represents all possible solutions to a recurrence relation before initial conditions are applied. For the recurrence relation \( a_n = -5a_{n-1} + 6a_{n-2} \), and after solving the characteristic equation, we found the roots to be \( r = 1 \) and \( r = -6 \).

These roots enable us to construct the general solution: \( a_n = A(1)^n + B(-6)^n \), where \( A \) and \( B \) are arbitrary constants. The terms \( (1)^n \) and \( (-6)^n \) correspond to the characteristic roots and illustrate the growth or decay pattern of the sequence. The coefficients \( A \) and \( B \) are determined by applying the initial conditions, hence shifting the general form to suit our specific sequence.
System of Equations
A system of equations emerges when we apply initial conditions to the general solution. By substituting \( a_0 = 5 \) and \( a_1 = 19 \) into the general solution, we derive two equations: \( A + B = 5 \) and \( A - 6B = 19 \).

These equations work together to lock in the values of \( A \) and \( B \). Solving this system involves a step-by-step algebraic process: we start by eliminating one variable, typically through substitution or elimination, simplifying the computation.

For instance, here by subtracting the first from the second equation, we simplified the system and found \( B = -2 \). Using this in the first equation, we quickly derived \( A = 7 \). Finally, replacing \( A \) and \( B \) back into the general solution, we completed the process, obtaining the particular solution, which describes our original sequence.

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Most popular questions from this chapter

Consider the sequence defined recursively by \(a_{1}=1\) and for \(n>1, a_{n}=\sum_{i=1}^{n-1} a_{i} .\) Write down the first six terms of this sequence, guess a formula for \(a_{n}\) valid for \(n \geq 2\), and prove your answer.

(a) [BB] Find, to four decimal places, the 129 th term of the geometric sequence which begins \(-0.00001240,0.00001364 .\) (b) \([\mathrm{BB}]\) Find the approximate sum of the first 129 terms of the sequence in (a).

An \(n\) -sided polygon (commonly shortened to \(n\) -gon) is a closed planar figure bounded by \(n\) straight sides no two of which intersect unless they are adjacent, in which case they intersect just at a vertex. Thus, a 3 -gon is just a triangle, a 4 -gon is a quadrilateral, a 5 -gon is a pentagon, and so on. An \(n\) -gon is convex if the line joining any pair of nonadjacent vertices lies entirely within the figure. A rectangle, for example, is convex. Prove that the sum of the interior angles of a convex \(n\) -gon is \((n-2) 180^{\circ}\) for all \(n \geq 3 .\)

Consider the geometric sequence with first term 59,049 and common ratio \(-\frac{1}{3}\). (a) Find the first ten terms and the 33 rd term of this sequence. (b) Find the sum of the first 12 terms.

For \(n \geq 1\), let \(b_{n}\) denote the number of ways to express \(n\) as the sum of 1 's and 2 's, taking order into account. Thus, \(b_{4}=5\) because \(4=1+1+1+1=2+2=\) \(2+1+1=1+1+2=1+2+1\). (a) Find the first five terms of the sequence \(\left\\{b_{n}\right\\}\). (b) Find a recursive definition for \(b_{n}\) and identify this sequence.

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