Question

Consider the following recurrence relation and initial conditions. tk=10tₖ₋₁−25tₖ₋₂ for each integer k≥2t₀=1,t₁=5 What value of t is a solution to this equation? t= It follows from part (a) and the roots theorem that for some constants C and D, the terms of t₀,t₁,t₂… satisfy the equation tn= for every integer n≥0 Solve for C and D by setting up a system of two equations in two unknowns using the facts that t0=1 and t₁=5. The result is that tn= for every integer n≥0.

Answer 1

To solve for C and D in the given recurrence relation, we substitute the initial conditions and set up a system of equations. By solving the system, we find that C=0 and D=1. Therefore, the general solution for tn is (-25)^n.

Step-by-step explanation:

To solve for C and D, we use the initial conditions t₀=1 and t₁=5 and substitute them into the given recurrence relation. We get:

t₀ = 10t₋₁ - 25t₋₂ = 10(5) - 25(1) = 50 - 25 = 25

t₁ = 10t₀ - 25t₋₁ = 10(25) - 25(5) = 250 - 125 = 125

Now we have the system of equations:

C + D = 1

10C + 5D = 5

Multiplying the first equation by 10, we get:

10C + 10D = 10

Subtracting the second equation from this, we eliminate the variable C:

10D - 5D = 10 - 5

5D = 5

D = 1

Substituting D=1 into the first equation, we find:

C + 1 = 1

C = 0

Therefore, the values of C and D are 0 and 1, respectively.

So the general solution for tn is:

tn = 0(10)^n + 1(-25)^n = (-25)^n