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Egyptian Fractions

Stage: 3 Challenge Level: Challenge Level:2 Challenge Level:2

This was a tough one, well done to lots of you who sent in lots of different examples, but only Rosie gave a reason and general rule for her findings:

I used Keep It Simple first to find that all unit fractions can be expressed as the sum of two other unit fractions like this,
$\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n(n+1)}$
I noticed that if $n$ was odd, then the numbers $n+1$ and $n(n+1)$ were both even. So if $n$ is odd, say for $m$ a positive integer, $n=2m+1$.
$\frac{2}{n}=\frac{2}{n+1}+\frac{2}{n(n+1)}$
$\frac{2}{n}=\frac{2}{2m+1}=\frac{2}{2m+2}+\frac{2}{(2m+1)(2m+2)}$
$\frac{2}{n}=\frac{2}{2m+1}=\frac{1}{m+1}+\frac{1}{(2m+1)(m+1)}$
Which are unit fractions.

Now if $n$ is even, then $n+1$ is odd, and so will not cancel like above. However, as $n$ is even than half of it is still a whole number. So if $n=2p$, $\frac{2}{n}$ is going to cancel to $\frac{1}{p}$, and we know that $\frac{1}{p}=\frac{1}{p+1}+\frac{1}{p(p+1)}$ from above.
So
$\frac{2}{n}=\frac{1}{p+1}+\frac{1}{p(p+1)}$
which we can write in terms of $n$
$\frac{2}{n}=\frac{2}{n+2}+\frac{4}{n(n+2)}$
These are unit fractions, and so we're done.

Great, can anyone use this to find $\frac{3}{n},\frac{4}{n},\frac{5}{n}$ and so on?



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