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How can I rewrite this equation:

$ \hspace{2in} V_0=\frac{E_1+B_0-B_1}{(1+r)^1} + \frac{E_2+B_1-B_2}{(1+r)^2} + \frac{E_3+B_2-B_3}{1+r)^3} + \dots $

into:

$ \hspace{2in} V_0=B_0+ \frac{E_1-rB_0}{(1+r)^1} + \frac{E_2-rB_1}{(1+r)^2} + \frac{E_3-rB_2}{(1+r)^3} + \dots $

My textbook simply states that this can be done without providing any hints.

I would also appreciate if someone could provide some resources (preferably online) on how to do this sort of algebraic manipulation. I had referred to my textbooks and could not find any hints on how to do this sort of algebraic manipulation.

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1 Answer 1

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Look at each E and B term on the right side. For example, in the top equation you have $\frac{B_0}{1+r}$. In the bottom equation you have $B_0-\frac{rB_0}{1+r}$. Can you see that these are equal? For $B_1$ you start with $\frac{-B_1}{1+r}+\frac{B_1}{(1+r)^2}$ and end with $\frac{-rB_1}{(1+r)^2}$. So $\frac{-B_1}{1+r}+\frac{B_1}{(1+r)^2}=\frac{-B_1(1+r)+B_1}{(1+r)^2}=\frac{-rB_1}{(1+r)^2}$. The rest of the B's are similar. The E's don't change at all.

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    $\begingroup$ And if you - as it happened to me - just don't see how to get from $\frac{B_0}{1+r}$ to $B_0 + \frac{-rB_0}{1+r}$ ... just try to start from the latter, it will be obvious then ;) $\endgroup$
    – Leonidas
    Commented Feb 1, 2011 at 4:53
  • $\begingroup$ Milikan: Thank you. I just have a quick question on the strategy to tackle this sort of algebra. Given $\frac{E_1+B_0-B_1}{(1+r)^1}$, how did you know to just look at $\frac{B_0}{1+r}$ and not other forms like $\frac{E_1+B_0}{1+r}$,etc? $\endgroup$
    – Sara
    Commented Feb 1, 2011 at 6:05
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    $\begingroup$ @Sara: Looking at the expression, nothing happened to the denominators or the E's, but all the B's after $B_0$ shifted one slot to the right. So I focused on the B's $\endgroup$ Commented Feb 1, 2011 at 13:25

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