How did the solution to this system of equations get a power of n?

Question

I have been reading up on how to solve problems relating to ideal gases. In a certain example problem in the book, Questions and Problems in school Physics by Tarasov and Tarasova, a system of equations were derived and was solved as shown below:

I understand how they got each of those equations but I can't understand how they solved them all(especially how there is suddenly a power of n in the solution). Help?

PS:- I felt this would be more appropriate in math.stackexchange rather than physics.stackexchange since it is the math that I don't seem to get here. I hope I am not wrong.

Answer 1

Note that the $p_{2}$ on the right of the second line is the same $p_2$ as on the left of the third line. Similarly with all others that appear.

So, using the final line and dividing each line by $(V+v)$ we have:

$$\begin{array}{l&l}p_n &= p_{n-1}(\frac{V}{V+v})\\ & = (p_{n-2}(\frac{V}{V+v}))(\frac{V}{V+v})\\ & = ((p_{n-3}(\frac{V}{V+v}))(\frac{V}{V+v}))(\frac{V}{V+v})\\ &\vdots\\ &=p_0 (\frac{V}{V+v})(\frac{V}{V+v})\cdots (\frac{V}{V+v})\\ &=p_0 (\frac{V}{V+v})^n \end{array}$$

This can be made more formal using an induction proof.

Answer 2

First equation implies $$p_1 = p_0 x, \text{ with } x = \frac{V}{V+v}.$$

Use second equation to show $$ p_2 = p_1 x = (p_0 x) x = p_0 x^2 $$ and prove the rest by induction with a similar argument.

Answer 3

Multiply all equations together and get $$ p_0 p_1 \cdots p_{n-1} V^n = p_1 \cdots p_{n} (V+v)^n $$ Cancel $p_1 \cdots p_{n-1}$ and get $$ p_0 V^n = p_{n} (V+v)^n $$ and so $$ p_{n} = p_0 \frac{V^n}{(V+v)^n} = p_0 \left(\frac{V}{V+v}\right)^n $$

Answer 4

With basic algebra

$$p_n = \frac{V}{V+v}p_{n-1}$$

Which means that to get to the next term you multiply by:

$$\frac{V}{V+v}$$

If you multiply more than once you have a power.