Question

Miranda arranges some rows of dominoes so that after she knocks over the first one, each domino knocks over two more dominoes when it falls. If there are ten rows, how many dominoes does Miranda use?

Answer 1

`2^10-1 = 1023`

Explanation:

The sum we are looking for is:

`1+2+4+...+2^9 = sum_(k=1)^10 2^(k-1)`

Note that:

`sum_(k=1)^n 2^(k-1) = (2-1)sum_(k=1)^n 2^(k-1)`

`color(white)(sum_(k=1)^n 2^(k-1)) = 2sum_(k=1)^n 2^(k-1) - sum_(k=1)^n 2^(k-1)`

`color(white)(sum_(k=1)^n 2^(k-1)) = sum_(k=2)^(n+1) 2^(k-1) - sum_(k=1)^n 2^(k-1)`

`color(white)(sum_(k=1)^n 2^(k-1)) = 2^n + color(red)(cancel(color(black)(sum_(k=2)^n 2^(k-1)))) - color(red)(cancel(color(black)(sum_(k=2)^n 2^(k-1)))) - 1`

`color(white)(sum_(k=1)^n 2^(k-1)) = 2^n - 1`

So we find:

`sum_(k=1)^10 2^(k-1) = 2^10-1 = 1023`

`color(white)()`
Footnote

Is this actually possible to set up physically?

Suppose our dominos are `2xx1` rectangles, say `2xx1` inches.

The maximum distance of any domino in the tenth row from the first is `9xx2=18` inches.

The tenth row contains `2^9 = 512` dominos.

So if they are standing up side to side, they span a total of `512` inches.

If that is a straight line, then the distance between the dominoes at the two ends is `510` inches (being the sum of the other `2^9-2` dominoes.

If it were more like a circle, then the circumference would be `512` inches and radius `512/(2pi) ~~ 81.5` inches.

How about the surface of a sphere?

The total surface area of one side of `512` dominoes of size `2xx1` inches would be `1024` square inches.

The surface area of a sphere is `4pir^2`. So a sphere of surface area `1024` square inches would have a radius:

`sqrt(1024/(4pi)) ~~ 9` inches.

That looks like it might be possible, but it is tricky to visualise an actual arrangement that would work.

`color(white)()`
Another possibility is somewhat easier:

• Make the first domino enormous.

• Make the dominos in each subsequent row half the size of the previous row.