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Given $$ P = \frac{A}{1+x} + \frac{A}{(1+x)^2} + \dots + \frac{A}{(1+x)^{N-1}} + \frac{A}{(1+x)^{N}}, $$

how can I arrive at the textbook's expression $$ P = A\left( \frac{1 - \frac{1}{(1+x)^N}}{x}\right) $$

Attempt at solution

This seems like a divergent geometrical series, I know that the partial sum $$ S_n = \frac{1-r^N}{1 - r}, $$ where $r$ is the common ratio $1 / (1+x)$, but when I write down $S_n$, I find an extra $1+x$ factor $$ P = A\frac{S_n}{1+x}. $$ It led me to think that maybe because in the definition of $S_n$ for an infinite geometrical series we start at $x^0$, i.e., $a + ax + ax^2 + \dots$, and in this problem we start at $ax + ax^2 + \dots$, then there is some term missing around, but I can't seem to make it work.

Can anyone point out what I am missing ?

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  • $\begingroup$ $a+ar+ar^2+\cdots +ar^{n-1} = a\frac{1-r^n}{1-r}$ so $ar+ar^2+\cdots +ar^{n-1}+ar^{n} = ar \frac{1-r^n}{1-r}$ $\endgroup$
    – Henry
    Jun 24 at 15:23
  • $\begingroup$ You may also pick common ratio $=(1+x)$ and the first term $=\frac{A}{(1+x)^N}$, then the denominator $x$ in the textbook answer becomes obvious. $\endgroup$
    – peterwhy
    Jun 24 at 15:34

2 Answers 2

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Here, we have $r=\frac{1}{x+1}$ is the common ratio. Next, we have \begin{align} P&:=Ar+Ar^2+\cdots +Ar^N\\ &=Ar(1+r+\cdots +r^{N-1})\\ &=Ar\left(\frac{1-r^{(N-1)+1}}{1-r}\right)\\ &=Ar\frac{1-r^N}{1-r}. \end{align} Plugging in $r=\frac{1}{x+1}$ gives the desired answer. The simple trick here to deal with the fact that the leading term is $Ar$ rather than 1, is to just factor out that 'troublesome' term.

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$ P = \frac{A}{1+x} + \frac{A}{(1+x)^2} + \dots + \frac{A}{(1+x)^{N-1}} + \frac{A}{(1+x)^{N}}$

$=\frac{A}{1+x}(1 + \frac{A}{1+x} + \frac{A}{(1+x)^2} + \dots + \frac{A}{(1+x)^{N-2}} + \frac{A}{(1+x)^{N-1}})$

$=\frac{A}{1+x}(\frac{1-\frac{1}{(1+x)^N}}{1-\frac{1}{1+x}})$

$= \frac{A}{1+x}(\frac{1-\frac{1}{(1+x)^N}}{\frac{x}{1+x}})$

$=A(\frac{1-\frac{1}{(1+x)^N}}{x})$

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