# tricky fractions

Use this series of fractions to find the value of x.

Find $x$, where $$\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\dots-\frac{1}{1024}=\frac{x}{1024}$$

*This problem is adapted from the World Mathematics Championships*

**Anwser**: 341

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**Adding the terms in pairs**

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

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

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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

$\dfrac{256+64+16+4+1}{1024}=\dfrac{341}{1024}$

**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

\end{align}$$