1. Solve the recurrence relation a(0)=a, 4a(n+1)=5a(n)?

Question

I need a solution, not only answer

Answer 1

a[n+1]= (5/4) a[n]
so a[n] = (5/4)a[n-1] = ... = (5/4)^n a[0] = a (5/4)^n

In this case you can derive the solution directly. But in general there are not the same general solution methods for finite difference equations as there are for differential equations, and you have to build a knowledge base of known patterns and recognise them.

a[n+1] = b a[n] will always have the form a[n]= Ab^n

Answer 2

So,

a[n+1] = 5/4 a[n]
= (5/4)^2 a[n-1]
= ....
= (5/4)^(n+1) a[0]

Applying the initial condition, you get
a[n+1] = a*(5/4)^(n+1).

Or, as it would more likely be written,

a[n] = a*(5/4)^n.

Answer 3

5a(n) = 4a(n + 1)

a(n + 1) = 5/4 * a(n)

a(n) = (5/4)^n * a

For a = 0:
a(0)=(5/4)^0 * a = a

Let's prove for n+1:

5a(n) = 5 * (5/4)^n * a= 4 * 5/4 * (5/4)^n * a = 4a*(5/4)^(n + 1) =
= 4a(n + 1)