The tutor shows how to find the general solution to the recurrence relation from last post.

In my previous post I posed the recurrence relation

2a_n+1 – a_n = 3, a₀=12

then generated the first few terms:

The general solution is a formula that, when n is inputted, yields a_n.

Example: Find the general solution to 2a_n+1 – a_n = 3, a₀=12

Solution:

Grimaldi reveals that for an equation of the form

a_n+1 + ra_n = k

the solution will take the form

a_n = Arⁿ + B

where A, B are constants that need to be solved.

To bring our equation to that form, we divide both sides by two:

a_n+1 – 0.5a_n = 1.5

The general solution will have the form

a_n = A(0.5)ⁿ + B

In our case, a₀ = 12:

a₀ = 12 = A(0.5)⁰ + B

so that

12 = A + B

Solving for B we get

12 – A = B

Then, for a₁ = 7.5, we get

7.5 = A(0.5)¹ + B

which means

7.5 = A(0.5) + B

Recalling 12 – A = B, we can rewrite:

7.5 = A(0.5) + 12 – A

which gives

7.5-12 = -0.5A

then

-4.5 = -0.5A

Next, dividing both sides by -0.5, we arrive at

9 = A

Then

12 – A = B

becomes

12 – 9 = B

so that

B = 3

Apparently, the general solution to 2a_n+1 – a_n = 3, a₀=12, is a_n = 9(0.5)ⁿ + 3

We seek to confirm the solution: From the table, a₄ = 3.5625. Plugging 4 in for n in our general solution gives

a₄ = 9(0.5)⁴ + 3 = 0.5625 + 3 = 3.5625

HTH:

Source:

Grimaldi, Ralph P. Discrete and Combinatorial Mathematics. Don Mills: Addison-
  Wesley, 1994.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.