91Ó°ÊÓ

In solving recurrence relations, one critical step is finding the characteristic equation. For our relation, \(a_{n+1} = 7a_n - 10a_{n-1}\), the goal is to convert it into something more manageable. We start by framing it as a characteristic equation. This involves assuming a potential solution in the form of \(a_n = r^n\). When substituted back into our recurrence relation, it results in an equation that equates powers of \(r\). Here, it simplifies into the quadratic equation \(r^2 = 7r - 10\). Rearranging terms gives us the characteristic polynomial: \[ r^2 - 7r + 10 = 0 \]. This quadratic equation holds the key to unlocking the general solution to the recurrence relation.

With the characteristic equation \(r^2 - 7r + 10 = 0\) derived, the next step is to solve it. This equation is quadratic, and we solve it by factoring. Factoring results in \((r - 5)(r - 2) = 0\), which gives us the roots \(r = 5\) and \(r = 2\). These roots are fundamental in constructing the general solution.

The general solution of our recurrence relation is thus a linear combination of the terms found by these roots. Specifically, it is given by: \[ a_n = A \, 5^n + B \, 2^n \]. Here, \(A\) and \(B\) are constants that we will find using the initial conditions provided. This formula allows us to express \(a_n\) in terms of the powers of the roots, reflecting the behavior dictated by the recurrence relation.

Initial conditions are vital for pinning down the exact form of our solution, especially when constants like \(A\) and \(B\) in our general solution are yet to be determined. For our problem, we have \(a_1 = 10\) and \(a_2 = 29\). These values help us create a system of equations tied to our general solution.

By substituting \(n = 1\) into the general solution equation, we get: \(5A + 2B = 10\). Similarly, substituting \(n = 2\) provides us with another equation: \(25A + 4B = 29\). These two equations represent the backbone of our solution refinement, linking the recurrence sequence to specific initial values.

The equations formed from our initial conditions \((5A + 2B = 10)\) and \((25A + 4B = 29)\) collectively form a system. This system needs to be solved for the coefficients \(A\) and \(B\). Solving systems of equations is a common method in mathematics, wherein we find the values of variables that satisfy all equations simultaneously.

Here's how it's done:

• First, we express \(A\) in terms of \(B\) from the first equation: \(A = (10 - 2B)/5\).
• Next, we substitute this expression for \(A\) into the second equation.
• On simplifying, we isolate \(B\) and find \(B = 3.5\).
• Finally, substituting \(B\)'s value back to find \(A\) gives \(A = 0.6\).

By solving these equations, we determine the values of \(A\) and \(B\), allowing us to finalize the specific solution for our recurrence relation: \(a_n = 0.6 \, 5^n + 3.5 \, 2^n\). This highlights how establishing and solving a system of equations is crucial in particularizing the general solution to our recurrence relations based on given initial conditions.