I am doing a question and instead of going through phases solving the question I was wondering if I could do it all in one with a short equation.

The question is about compound interest finding the future value. The number $2$ is used for ease of writing but it would be the interest. $500$ is the amount being added every six months.

So I would like to do something like:

$A = 500(2^{20}+\dots+2^3+2^2+2^1+2^0)$

The exponent would be the number of conversion periods.

We have learned to do it step by step with: $A = P(1+i)^n$ I would have to calculate the first six months then add the $500$ and continue doing this until I reach ten years.

This way seems inefficient so I was wondering if there was an easier way. :)


  • 2
    $\begingroup$ This is a geometric series. The sum can be represented nicely for any base, not just $2$ $\endgroup$ – Ross Millikan May 24 '14 at 21:43

Well, there is some notation:

$$2^n + 2^{n-1} + 2^{n-2} + ... + 2 + 1 = \sum_{k=0}^n 2^k$$

It's a different question altogether whether an explicit formula can be given to evaluate this sum. As it turns out, one does exist. See Nishant's answer (+1) and this Wikipedia article.


$a^0+a^1+\cdots+a^n=\frac{a^{n+1}-1}{a-1}$, assuming $a\neq 1$. If $a=1$, then the sum is obviously $n+1$.


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