Solve a 2nd order homogeneous difference equation

Solve a 2nd order homogeneous difference equation?
Solve the difference equation
an+2 = 2 an+1 - an

given a0= 6 and a1= 7

an= ?

so the answer is 6+n...but i have no idea how to get to that.

Say a_n = r^n. Then we have:
r^(n+2) = 2r^(n+1) - r^(n)

Divide everything by r^(n) and this yields:
r^(2) = 2r - 1

Solve this like a quadratic equation:
r^2 - 2r + 1 = 0
(r - 1)^2 = 0
r = 1

In the case that r is only one real solution, you will have then have something in the form a_n = c₁(r₀^n) + c₂n(r₀^n). Just in case your curious, if you have two real solutions then it will be in the form a_n = c₁(r₁^n) + c₂(r₂^n).

So for yours we found that r₀ = 0, which means we have:
a_n = c₁(1)^n + c₂n(1)^n

Of course, (1)^n = 1, so:
a_n = c₁ + c₂n

Use the initial conditions:
a₀ = c₁ + c₂(0) = 6 ==> c₁ = 6
a₁ = c₁ + c₂(1) = 7 ==> 6 + c₂ = 7 ==> c₂ = 1

Thus we get:
a_n = 6 + n

To actually prove a formula for an requires a theoretical mathematical proof by a method known as Strong Induction, which I'm not sure you've covered yet.

Thus, the only other way to look at this is to calculate some values and see if we detect a trend in the data.

a0 = 6
a1 = 7
Then a2 = 2a1 - a0 = 14 - 6 = 8
So when n = 0, we get a2 = 8

a3 = 2a2 - a1 so a3 = 16 - 7 = 9
a4 = 2a3 - a2 so a4 = 18 - 8 = 10

etc.

so a0 = 6
a1 = 7
a2 = 8
a3 = 9
a4 = 10

We can see that if an = 6 + n, that this works.
a0 = 6 + 0 = 6
a1 = 6 + 1 = 7
a2 = 6 + 2 = 8
a3 = 6 + 3 = 9
a4 = 6 + 4 = 10
etc.

If you have covered Strong Induction in a theoretical math class, email me and I'll be happy to show you how to prove that this is the formula using that method.

a[n] = 6 + n