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How do you go from

$$q_1 = rq_1 + pq_2$$ $$q_2 = rq_2 + pq_3$$ $$q_3 = rq_3 + p $$

to get

$$q_1 = \left( \frac{p}{1 - r} \right) ^3 $$

Because of that extra $q_3$, I keep getting lost in what I'm doing. Plus I can't get the $(1 - r)^3$.

How do you do this?

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3 Answers

up vote 3 down vote accepted

$$q_3 = rq_3 + p $$

gives

$$q_3 = p/(1-r)$$

so

$$q_2 = rq_2 + p^2/(1-r)$$

which gives

$$q_2 = p^2/(1-r)^2$$

so

$$q_1 = rq_1 + p^3/(1-r)^2$$

which gives

$$q_1 = p^3/(1-r)^3$$

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The easiest way to see what's going on here, is probably to start rewriting the system of three equations as $(1-r)q_1 = pq_2$, $(1-r)q_2 = pq_3$, $(1-r)q_3 = p$. Then you divide everything by $(1-r)$ and just plug in the different $q_i$, like @user58512 did. –  HSN Jan 20 '13 at 18:30
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Note that the equations can be rearranged to get \begin{align} (1-r)q_1 & = pq_2 & \implies q_1 = \dfrac{pq_2}{1-r} & (\star)\\ (1-r)q_2 & = pq_3 & \implies q_2 = \dfrac{pq_3}{1-r} & (\perp)\\ (1-r)q_3 & = p & \implies q_3 = \dfrac{p}{1-r} & (\dagger)\\ \end{align} Hence, from $(\dagger)$ we get that $$q_3 = \dfrac{p}{1-r}$$ Plugging the above in $(\perp)$, we get that $$q_2 = \dfrac{p^2}{(1-r)^2}$$ and plugging the above in $(\star)$, we get that $$q_1 = \dfrac{p^3}{(1-r)^3}$$

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Note that the equations are $$(1-r)q_1=pq_2\qquad\ldots(1)$$ $$(1-r)q_2=pq_3\qquad\ldots(2)$$ $$(1-r)q_3=p\qquad\ldots(3)$$ Now put the value of $q_3$ from (3) into (2) to get $q_2$ and finally put the value of $q_2$ into (1).

OR Just multiply the three equations....

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