One way is to work it out. Write down a few terms and see what emerges in your mind.

3

-2.6

-7.08

...

Okay, that is totally not what I meant.

n=1) 3

n=2) 3*(4/5) - 5

n=3) (4/5)[3*(4/5) - 5] - 5 = 3*(4/5)^2 - 5(4/5) - 5

I'm beginning to see it. Are you? Let's just guess what the next one will be.

n=4) 3*(4/5)^3 - 5*(4/5)^2 - 5(4/5) - 5

And the nth one? (Feel free to prove this by induction.)

n) 3*(4/5)^(n-1) - 5(4/5)^(n-2) - ... - 5*(4/5)^2 - 5(4/5) - 5

Make a small adjustment and add them up.

n) 8*(4/5)^(n-1) - 5*(4/5)^(n-1) - 5(4/5)^(n-2) - ... - 5*(4/5)^2 - 5(4/5) - 5

Some algebra

n) 8*(4/5)^(n-1) - 5*[(4/5)^(n-1) + (4/5)^(n-2) + ... + (4/5)^2 + (4/5) + 5]

A little more algebra

n) 8*(4/5)^(n-1) - 5*[(1 - (4/5)^n)/(1-(4/5))

Simplify

n)

I think I'm done. Of course, you're not. You still have to prove the formula.

Note: There is a better way. It starts with simply assuming the form of the final result and solving for a few parameters. In this case, there is a hint. Rewrite the formula a little:

Compare this to the final result. From now on, with a first order recursion and a constant

You say to yourself, it must look like this.

Use your initial values to solve for D and you are rather done.