You may also like

Consecutive Numbers

An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.

Have You Got It?

Can you explain the strategy for winning this game with any target?

Counting Factors

Is there an efficient way to work out how many factors a large number has?

Tricky Fractions

Age 11 to 14 Short
Challenge Level

Anwser: 341

 Adding the terms in pairs

The first pair of terms add up to $\dfrac{1}{2}-\dfrac{1}{4}=\dfrac{1}{4}$

The next pair of terms add up to $\dfrac18-\dfrac1{16}=\dfrac{1}{16}$


The next pair of terms ($\frac1{32}-\frac1{64}$) add up to $\dfrac1{64}$

... the last pair of terms add up to $\frac1{1024}$

The whole sum is $\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}+\dfrac{1}{1024}$

Common denominator: $1024$

Each fraction is $4$ times smaller than the last so whole sum is equal to


Multiplying the sequence by 2
If $$\frac{x}{1024}=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\dots-\frac{1}{1024}$$then $$\begin{align}\frac{2x}{1024}=&2\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\dots-\frac{1}{1024}\right)\\=&1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\dots-\frac{1}{512}\end{align}
$$Notice that most of the terms in $\dfrac{2x}{1024}$ are the same as the terms in $\dfrac{x}{1024}$  $$\begin{align}\frac{2x}{1024}+\frac{x}{1024}=1-&\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\dots-\frac{1}{512}\\+&\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\dots-\frac{1}{1024}\\

\Rightarrow \frac{3x}{1024}=1+&\text{ }0\quad+\ \quad\dots\ \quad+\quad 0\text{ }- \frac{1}{1024}=\frac{1023}{1024}&\\
\therefore x=1023&\div3=341
You can find more short problems, arranged by curriculum topic, in our short problems collection.