I've been a bit stumped on how to write an equation on increasing costs and I'm wondering how I'd go about it. Any help is very much appreciated.
Say I'm a real estate speculator.
$X$ = How many houses I own
$Y$ = How many houses I want to own
$Z$ = Total cost
(1st part) In America I want to buy houses. The cost of houses in America are $ \$ 1000$ each.
The equation for the above would be $(Y-X)⋅1000=Z$ e.g. If I want to own $5$ houses and currently own $3$ it's going to cost me $(5-3)⋅1000=\bf2000$, easy. Although the second part is not so easy for me.
(2nd part) In Europe the cost of houses are $1000$ plus a $500$ tax per house I already own e.g. my 1st home purchase costs me $1000$, 2nd $1500$, 3rd $2000$, 4th $3500$ etc.
If I currently owned $0$ houses and wanted to own $10$ how would I write an equation to figure out the total cost?
As a followup question if I currently owned $4$ houses and wanted to own $7$ could I use the same equation to figure out the total cost?
Thank you very much for reading =)
Imagine you own $X$ houses already and call $P_1$ the price of the first house you buy, $P_2$ the price of the second house you buy ...
\begin{eqnarray} P_1 &=& 1000 + 500 X \\ P_2 &=& 1000 + 500 (X+1)\\ &\vdots& \\ P_n &=& 1000 + 500(X + n - 1) \end{eqnarray}
That is, the price of the $n$-th house you buy will be
$$ P_n = 1000 + 500(X-1) + 500 n \tag{1} $$
The total cost of buying $Y-X$ is just
\begin{eqnarray} T &=& P_1 + \cdots + P_{Y-X} \\ &=& [1000 + 500(X-1)](Y-X) + 500 \color{red}{[1 + 2 + \cdots + (Y-X)]} \\ &=& [1000 + 500(X-1)](Y-X) + 500 \color{red}{\frac{(Y-X)(Y-X + 1)}{2}} \tag{2} \end{eqnarray}
Where I've used
$$ 1 + 2 + \cdots + n = \frac{n(n+1)}{2} \tag{3} $$