### Converging Product

In the limit you get the sum of an infinite geometric series. What about an infinite product (1+x)(1+x^2)(1+x^4)... ?

A circle is inscribed in an equilateral triangle. Smaller circles touch it and the sides of the triangle, the process continuing indefinitely. What is the sum of the areas of all the circles?

### Binary Squares

If a number N is expressed in binary by using only 'ones,' what can you say about its square (in binary)?

# Clickety Click and All the Sixes

##### Stage: 5 Challenge Level:

This caused a lot of thinking and Alex of Madras College gave the following proof.

First consider $S_n = 1 + 11 + 111 + 1111 + \cdots$ to $n$ terms.

Each individual term can be written and summed as a geometric series, for example $$1111 = 1 + 10 + 100 + 1000 = \frac{10^4-1}{10 - 1}$$ Hence $$S_n= \frac{10^1 - 1}{9} + \frac{10^2 - 1}{9} + \frac{10^3 - 1}{9} + \frac{10^4 - 1}{9} + ... + \frac{10^n - 1}{9}$$ $$= \frac{10 + 10^2 + 10^3 + 10^4 + ... +10^n }{9} - \frac{n}{9}$$ $$= \frac{10^{n+1}- 10}{81} - \frac{n}{9}$$ $$= \frac{10^{n+1}- 10 - 9n}{81}$$ So $6 + 66 + 666 + 6666 \cdots$ to $n$ terms is: $$6( 1 + 11 + 111 + 1111 + ... ) = \frac{2}{3}\Big[ \frac{10(10^n - 1)}{9}- n \Big]$$

Remarkably this result was submitted on the 1st of March by Chong Wenhao Edmund from Singapore, the earliest solution! No doubt the time difference helped.