Alan

You could also look at it this way:

You can see that this will always end with 10/18, i.e. there will always be a remainder of 10.

\(a+ar+ar^2+...=\frac{a}{1-r}\) as long as r<1

\(a^3+a^3(r^3)+a^3(r^3)^2+...=\frac{a^3}{1-r^3}\)  as long as r^3<1

So:  

\(\frac{a}{1-r}=4\)

and

\(\frac{a^3}{1-r^3}=10\)

Can you take it from here?

The other condition seems to be that you specify in advance the values of x, y and z.  It wasn't clear from your original question that that was what was intended!

There are an infinite number of solutions to the question you asked - see below.  Perhaps you meant to add another condition.

Like so:

Use \(\lim_{n\rightarrow\infty}Summation\)

I'll leave you to write the whole expression in LaTeX.

I couldn't see a neat way of doing this analytically, so I used a numerical approach:

As follows:

@hairyberry  You need to check \(f(4\sqrt5)\).  It isn't 70.

"I wasn't really sure if you could separate the magnitude of z+w to the sum of two magnitudes."

You can't in general!

Any permutation starting 1, 5 , ...    4! of them 

Any permuttion starting 1, 6, ...       4! of them

Any permutation ending ..., 5, 1      4! of them

Any permutation ending ..., 6, 1      4! of them

Any permutation starting 5, 1, 6, ...  3! of them

Any permutation starting 6, 1, 5, ...  3! of them

etc. ...

The limit of the argument is zero, but that doesn't make the sum zero.

Use the following:

As follows:

@Padewolofoofy:  You forgot to divide by 2^2