We will begin to calculate, to see what is going on.
We have $P_0=40$. Therefore, by the given recurrence, $P_1=(1.75)P_0=(1.75)(40)$.
Important: Don't plug $(1.75)(40)$ into a calculator! In fact, to prevent the temptation, I will write instead $P_1=(1.75)P_0$. We want to be always aware of the structure of our answer, and plugging numbers into a calculator tends to hide structure.
What is $P_2$? by the recurrence, it is $(1.75)P_1$. But $P_1=(1.75)P_0$, so
$P_2=(1.75)(1.75)P_0$. For convenience write instead $P_2=(1.75)^2P_0$.
What is $P_3$? By the recurrence, $P_3=(1.75)P_2$. But from our expression for $P_2$, we get $P_3=(1.75)(1.75)^2P_0$, which is $(1.75)^3P_0$.
Maybe compute a bit further. Or not. It is I think clear that the same reasoning will give $P_4=(1.75)^4P_0$, and in general $P_n=(1.75)^nP_0$ for any non-negative integer $n$. So we conclude that
$$P_n=((1.75)^n)(40).$$
Now let's look at your $Y=5.4x -27$. Presumably the intent is that $x$ is a non-negative integer. In order to bring out the analogies with the first problem, I will write $n$ instead of $x$. So $Y=5.4n-27$. Note that $Y$ depends on $n$. Bring out this dependence by writing instead
$$Y_n=5.4n -27.$$
Note that $Y_0=-27$. What is $Y_1$? It is $5.4$ more than $Y_0$. So $Y_1=5.4+Y_0$.
What is $Y_2$? From the formula, it is $(2)(5.4)-27$. Note that this is $5.4$ more than $Y_1$. And what is $Y_3$? It is $(3)(5.4)-27$, which is $5.4$ more than $Y_2$. It is clear that $Y_4$ is $5.4$ more than $Y_3$, and that in general $Y_n$ is $5.4$ more than $Y_{n-1}$.
We reach the conclusion that $Y_n=Y_{n-1}+5.4$, with $Y_0=-27$.
We could proceed slightly more abstractly. Note that $Y_n=5.4n-27$ and that $Y_{n-1}=5.4(n-1)-27$. Subtract. After a bit of algebra, we get $Y_n-Y_{n-1}=5.4$, or equivalently $Y_n=Y_{n-1}+5.4$.
Finally, we look at the equation that you wrote as
$$Y=3800(1.04)^x$$
Again, I imagine $x$ is supposed to be a non-negative integer. I will keep the $x$, though I am tempted to replace it by $n$. But definitely, "$Y$" is not good, we should write it either as $Y_x$ or $Y(x)$, to make explicit the dependence on $x$.
So we have
$$Y_x=3800(1.04)^x$$
Now look at my discussion of the first question you had. What was the answer to that? And what was the question? In our current problem, the question looks like the answer to the first problem we dealt with.
Or more directly, we have from the given equation that $Y_0=3800$. Can you see that to get the "next" $Y$, you multiply the current $Y$ by $1.04$? So $Y_x=(1.04)Y_{x-1}$.
The conventional way to get a closed form for that recurrence would be first to get rid of the constant term by subtracting the equation for $n-1$ from the one for $n$:
b(n) = b(n-2) + b(n-3) - b(n-5) + 1
- b(n-1) = b(n-3) + b(n-4) - b(n-6) + 1
-------------------------------------------------------------
b(n) = b(n-1) + b(n-2) - b(n-4) - b(n-5) + b(n-6)
You then need to find the roots of the characteristic polynomial of this equation,
$$t^6 - t^5 - t^4 + t^2 + t - 1 $$
I cheated and used Wolfram Alpha for that, but it can also be done by hand in this case because there are many $\pm1$ roots. It factors as
$$ (t^3-1)(t^2-1)(t-1) = (t+1)(t-1)^2(t^3-1) = (t+1)(t-1)^3(t-\zeta)(t-\zeta^2) $$
where $\zeta$ is a complex third root of unity. (That $(t^3-1)(t^2-1)(t-1)$ factorization sure looks like it has something to do with the 1,2,3 of your original problem statement, doesn't it?)
So the general solution of the recurrence is
$$ b(n) = c_1 + c_2n + c_3n^2 + c_4(-1)^n + c_5\zeta^n + c_6\zeta^{2n} $$
Since the sequence is obviously real, we must have $c_5=c_6$, so this becomes
$$ b(n) = c_1 + c_2n + c_3n^2 + c_4(-1)^n + c_5 \cdot\begin{cases} 2 & \text{when }3\mid n \\ -1 & \text{otherwise}\end{cases}$$
You will then need to find the $c_i$s by solving the system of linear equations you get by plugging in 5 known values of $b(n)$. Finally you can plug $n=2012$ and so forth into the formula.
Edit: Oops -- the appropriate condition for the result to be real is not $c_5=c_6$, but $c_5=\overline{c_6}$, and there can be an imaginary part in these constants. It is probably easiest then to subsume $c_1$, $c_5$ and $c_6$ into new (real) constants $k_0$, $k_1$, $k_2$, and write
$$ \tag{*} b(n) = c_2n + c_3n^2 + c_4(-1)^n + k_{n\,\bmod\,3}$$
Then six known values of $b(n)$ are necessary to find all of the constants -- and a good thing that, because five values would not have been enough to know from the sample values what the constant term we started out by eliminating from the recurrence was.
Even more edit: Note that the form of $(*)$ makes good geometric sense, since the original problem
let b(n) be the number of ways that you can write n as a sum using only the numbers 1, 2, and 3 where the order of the sum doesn’t matter.
simply asks for the number of nonnegative integer solutions to $2x+3y\le n$, or in other words the number of points with integral coordinates in a triangle bounded by the horizontal and vertical axes and a line with slope $-\frac32$. To a first approximation we'd expect this to be $\frac 12 (n/3)(n/2)$, so we ought to have $c_3=\frac{1}{12}$. The other terms are there just to correct for fencepost errors. It is to be expected that there are corrections with period 2 and 3 as the axis intercepts of the hypotenuse sweeps past lattice points. It may be surprising that there is no similar periodic variation in the coefficient of the linear term $c_2n$, but that is probably because $2$ and $3$ are coprime, so the density of lattice points on the hypotenuse is the same for all $n$. It would be a different matter if it had been "using only the numbers 1, 2 and 4", for example. End edits.
If simply finding the values of $b(2012)$ upto $b(2017)$ is all you want, it is probably less work just to program a computer to fill in a table of $b(n)$ upto that number, using your recursive equation to fill in each value based on the ones you have already computed.
Best Answer
If you define $g(n) = \frac{f(n)}{n}$ then your expression is $g(n) = g(n-1) + \frac{0.01}{n}$ and $g(0) = 0$. So $g(n) = 0.01\sum_{i=1}^n \frac{1}{n} = 0.01 h(n)$ where $h(n)$ is the $n^{\text{th}}$ harmonic number. Therefore $f(n) = 0.01n \cdot h(n)$. I think this is about as closed of a form as you're going to get.
Since the sequence $h(n)$ diverges, you will eventually get $h(n)\geq 100000$ so $g(n) \geq 1000$, which means $f(n)\geq 1000n$.