Tricky fractions

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 $$\frac{x}{1024}=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\cdots -\frac{1}{1024}$$then $$\begin{array}{rl}\frac{2x}{1024}=& 2\times (\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\cdots -\frac{1}{1024})\\ =& 1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots -\frac{1}{512}\end{array}$$Notice that most of the terms in $\dfrac{2x}{1024}$ are the same as the terms in $\dfrac{x}{1024}$  $$\begin{array}{rl}\frac{2x}{1024}+\frac{x}{1024}=1-& \frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots -\frac{1}{512}\\ +& \frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\cdots -\frac{1}{1024}\\ \Rightarrow \frac{3x}{1024}=1+& 0+\dots +0-\frac{1}{1024}=\frac{1023}{1024}& \\ \therefore x=1023& ÷3=341\end{array}$$