Copyright © University of Cambridge. All rights reserved.

## 'Clickety Click and All the Sixes' printed from http://nrich.maths.org/

*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.*