Dear Uncle Colin,

As a challenge, I need to work out $1024 - 512 + 256 - \dots + 1$, but I’m getting mixed up on the fractions and signs. Can you help?

• Colin, Help! Answer Largely Lacking for Evaluating Negative Geometric Expression

Hi, CHALLENGE, and thanks for your message!

I’ve got three methods for you: the standard method (which I imagine is what you’ve tried and got stuck with), and a couple of tricks that work here.

### The standard method

The standard way would be:

• $a = 1024$
• $r = -\frac{1}{2}$
• We need to do some work to get $n$

So:

• $1024 \left(-\frac{1}{2}\right)^{n-1} = 1$
• Multiply both sides by $(-2)^{n-1}$ to get $1024 = (-2)^{n-1}$
• Since $2^{10} = 1024$, $(-2)^{10}$ is also 1024 and $n=11$.

Then we can use the sum formula $S_n = \frac{a\left(1-r^n\right)}{1-r}$ - I’ve picked the version that’s this way round to make the minus signs easier to work with.

You could (and, some would argue, should) stick the values into the calculator; it’s good for your soul to work things out by hand, though.

• $S_{11} = \frac{1024\left(1 + \frac{1}{2048}\right)}{1 + \frac{1}{2}}$
• Double top and bottom to get $\frac{2048\left(1 + \frac{1}{2048}\right)}{3}$
• Expand the top to get $\frac{2049}{3}$, and you can work this out to be 683.

### Alternative 1

An alternative that works here is to notice that you can reverse the series and call it $1 - 2 + 4 - … - 512 + 1024$.

Here $a=1$, $r=-2$ and $n=11$ again, and the sum is $\frac{1(1 + 2048)}{1+2}$, which gives you the result a bit more directly.

### Alternative 2

In this example, it’s quite easy to pair off numbers.

$(1024 - 512) + (256 - 128) + (64-32) + (16-8)+ (4-2)+ 1 = (512 + 128 + \dots 2) + 1$. I’ve left the 1 out of the bracket at the end because it isn’t part of a pair, and therefore isn’t part of the series.

Now we have $a = 512$, $r=\frac{1}{4}$ and $n=5$, so the geometric part sums to $\frac{512\left(1 - \left(\frac{1}{4}\right)^5\right)}{1-\frac{1}{4}}$ (and we’ll need to add 1 at the end).

Multiply top and bottom by 4: $\frac{2048\left(1 - \frac{1}{1024}\right)}{3}$.That’s $\frac{2046}{3} = 682$, and when we add the 1 back on, we get the same 683 again.

Hope that helps!

- Uncle Colin