Turning 500, 1100, 1800, 2600, 3500, etc into 1, 2, 3, 4, 5, etc I would love to have a formula of the form $n = F(m)$ for this progression ...
n      m    
---------
1     500
2    1100
3    1800
4    2600
5    3500
...

where the second-differences in $m$ are constant, on up to $n = 100$.
 A: Divide the elements by $100$ to get the sequence $5,11,18,26,35,...$, which is
1  5
2  5+6
3  5+6+7
4  5+6+7+8
...

which is
1  1+2+3+4+5  -  10
2  1+2+3+4+5+6  -  10
3  1+2+3+4+5+6+7  -  10
...

But these sums are triangle numbers, so the terms are given by $\frac{(n+4)(n+5)}{2} -10$. Now multiply by $100$ to recover the original terms:
$$f(n) = 50(n+4)(n+5)-1000 = 50 n^2 + 450 n$$
which can then be inverted using the quadratic formula to obtain
$$n = \frac{-450 + \sqrt{450^2 + 4\cdot 50\cdot f(n)}} {2\cdot 50}.$$
A: The sequence $\{500,1100,1800,2600,3500\}$ is a quadratic progression, since it's first differences are increasing linearly. So if these are $\{b_0,b_1,b_2,b_3,b_4\}$, then we can write $$b_n= An^2+Bn+C$$ Clearly $C=500$. 
We also have (from $n=1$) that $600=A+B$ and (from $n=2$) that $1300=4A+2B$. From these two equations you can solve for $A$ and $B$. ($A=50, B=550$)
So then you know $$b_n= 50n^2+550n+500\text{.}$$ Use the quadratic formula to invert this formula, choosing the appropriate sign ($+$ or $-$) to reach your goal:  $$n=\frac{-550+\sqrt{550^2-200(500-b_n)}}{100}=\frac{-55+\sqrt{55^2-2(500-b_n)}}{10}=\frac{-55+\sqrt{2025+2b_n}}{10}$$
Except I foolishly set this up to output $\{0,1,2,3,4\}$. So add $1$ to this formula.
A: Easier to go backwards: from $1$, get $500$; from $2$, get $1100$; from $3$, get $1800$; etc. You notice that the sequence $1, 2, 3,\dots$ has successive differences $1$ (constant), while the sequence $500,1100, 1800,\cdots$ has successive differences $600$, $700$, $800$, etc., linearly increasing. I’m not going to go into a discussion of finite differences, but:
You expect that the sequence of big numbers $N$ should depend quadratically on the sequence of little numbers $n$. So $N=an^2+bn+c$, for a good choice of $a$, $b$, and $c$. Pluggging in the first three values of $n$, you get three simultaneous equations for the three unknowns $a$, $b$, and $c$:
$$
\matrix{
a&+b&+c&=500\\
4a&+2b&+c&=1100\\
9a&+3b&+c&=1800\,.
}
$$
Next, subtract the top equation from each of the bottom two:
$$
\matrix{
3a&+b&=600\\
8a&+2b&=1300\,,
}
$$
and then subtract twice the top here from the bottom to get $2a=100$, $a=50$. then you see that $b=450$ and $c=0$. Your relation is $N=50n^2+450n$.
A: Let's first solve this problem backwards: let's say we want to find a sequence where $a_1 = 500, a_2 = 1100$ and so on.  With a little bit of work, we can find that $a_n = 50n^2 + 450 n$.  Let's say we know $a_n$ and want to know $n$; if we solve for $n$ is $0 = 50n^2 + 450 n - a_n$, we find that the function in question is $$n = \frac{-450 + \sqrt{(450^2) + 200\cdot a_n}}{100} $$ 
A: The relationship of your input in terms of your output can be explained as follows:
$$500x+100\sum_{i=0}^xi\,\mid\,\,x\in\Bbb N_0 $$
Explanation: Suppose for each output value exists $x \,\mid\,\,x\in\Bbb N_0$. Now, we say our input equals some value added to $500x$ (for convenience's sake). If $x=1$, I already get my desired input of $500(1)$. However, for $x=2$, my input value is $500(2)+100$, or $1100$. For $x=3$, I must add $300$ to $500(3)$, and for $x=4$, I must add $600$. The obvious relationship is that I am adding $100$ times the $x$th triangular number. To generate triangular numbers, I used a basic summation to generate desired triangular numbers.
I hope this helps!
