# Can someone point me in the direction of why $\sum\limits_{n=0}^{T-1}A(1+R)^n = A\frac{(1+R)^T-1}{R}$

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 $A\frac{(1+R)^T-1}{R}$?

• This is a geometric series, here's a link that explains the basic idea of how to arrive at the formula. – pjs36 Apr 1 '15 at 2:50
• @pjs36 Ah that helped a bunch, I understand now thanks! – Loocid Apr 1 '15 at 2:53

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: $$S= A\frac{(1+R)^T-1}{R}$$
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*}