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Before you ask, this isn't a homework question, I am just curious.

I was trying to derive an expression for compound interest with evenly spaced deposits.

I reached the point: $F = I(1+R)^T + \sum\limits_{n=0}^{T-1}A(1+R)^n$

I = Initial deposit

A = Annual deposit

R = Interest rate

T = Period

Obviously this isn't the most easy expression to work with so I looked up the correct answer which is apparently:

$F = I(1+R)^T + A\frac{(1+R)^T-1}{R}$

Apparently these expression are equivalent, so my question is, how do I get from:

$\sum\limits_{n=0}^{T-1}A(1+R)^n$ to $A\frac{(1+R)^T-1}{R}$?

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    $\begingroup$ This is a geometric series, here's a link that explains the basic idea of how to arrive at the formula. $\endgroup$ – pjs36 Apr 1 '15 at 2:50
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    $\begingroup$ @pjs36 Ah that helped a bunch, I understand now thanks! $\endgroup$ – Loocid Apr 1 '15 at 2:53
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this is a standard geometric series summation. let the first term be $a$ and the factor of increase be $x$. suppose there are $n-1$ increases, then: $$ S = a + ax + ax^2 + \cdots + ax^{n-1} $$ multiply by $x$ $$ Sx = ax + ax^2 + ax^3 + \cdots + ax^n $$ subtract the first from the second: $$ S(x-1) = ax^n -a $$ this is the trick! all terms cancel except the two extremes.

simplifying: $$ S = a\frac{x^n-1}{x-1} $$ since in your case we have $$ x=1+R \\ T=n $$ the formula becomes: $$ S= A\frac{(1+R)^T-1}{R} $$

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Your function is basically in the form of a geometric series, which we can write as $$ S = \sum_{n=0}^{N} ar^n $$ for first term $a$ and common ratio $r$. The easiest way to see how to compute the sum is to write it out longhand, multiply it by $r$, and then subtract: $$ \begin{align*} S &= a + ar + ar^2 + \dotsb + ar^n \\ rS &= \phantom{a+{}} ar+ ar^2 + \dotsb + ar^n + ar^{n+1} \\ (r-1)S &= -a + ar^{n+1} \\ S &= a\frac{r^{n+1}-1}{r-1} \end{align*}$$

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