### Golden Powers

You add 1 to the golden ratio to get its square. How do you find higher powers?

### 2^n -n Numbers

Yatir from Israel wrote this article on numbers that can be written as $2^n-n$ where n is a positive integer.

### Poly Fibs

A sequence of polynomials starts 0, 1 and each poly is given by combining the two polys in the sequence just before it. Investigate and prove results about the roots of the polys.

# Production Equation

##### Age 16 to 18 Challenge Level:

Simon from Elizabeth College, Guensey and Andrei from Tudor Vianu National College, Romania both solved this problem using geometric series. Here is Simon's solution:

The equation to show stock levels in week $n$ is: $$U_n = X + \left(1 - {p\over 100}\right)U_{n-1}.$$ Substituting for $U_{n-1}$ and simplifying:
\begin{eqnarray} U_n &=& X + \left(1 - {p\over 100}\right) \left(X + \left(1 - {p\over 100}\right) U_{n-2}\right) \\ &=& X + \left(1 - {p\over 100}\right)X + \left(1 - {p\over 100}\right)^2U_{n-2}. \end{eqnarray}
Originally there was no stock so the initial condition is $U_0 = 0$. To make the expressions easier to read we write $s= \left(1 - {p\over 100}\right)$.

Writing $U_n$ in terms of $U_0$: $$U_n = X(1 + s + s^2 + ... + s^{n-1})+ s^nU_0 = \sum_{r=0}^{n-1}Xs^r.$$ Summing this geometric series $$U_n = {X(1-s^n)\over 1 - s} = {100X\over p}\left(1 - \left(1 - {p\over 100 }\right)^n\right).$$ To find the limit of $U_n$ over a very long period of time, as $(1 - p/100)< 1$, we have $$\lim_{n\to \infty}\left(1- {p\over 100}\right)^n \to 0$$ and so $$\lim_{n\to \infty}U_n = {100X \over p}.$$ The company should ensure that they have enough warehouse space for ${100X\over p}$ items. However, the chances are that sales and production will vary during the year, and therefore they should ensure that they have a buffer zone so that they can have more stock if necessary. Also, they are unlikely to be in business for an infinite length of time and therefore should cater said limit to their needs based on the financial situation.