Wheat and Chessboard Problem

Classic Problem
The squares of a chessboard are numbered in some way, from $1$st to $64$th.

A grain of wheat is placed on the $1$st squares of the chessboard.

$2$ grains of wheat are placed on the $2$nd square.

$4$ grains of wheat are placed on the $3$rd square.

The sequence continues: on the $n$th square, twice as many grains of wheat are placed as there are on the $n - 1$th square.

How much wheat is there on the chessboard when the $64$th square has been filled?

Solution
There will be of the order of $1$ billion ($10^{12})$ cubic metres of grain on the chessboard.

Proof
Let $W$ be the total volume of grain on the chessboard.

Let $N$ be the number of grains on the chessboard

$N$ is governed by a geometric sequence of $64$ terms whose initial term is $1$ and whose common ratio is $2$.

From Sum of Geometric Sequence $N$ is given by:


 * $N = \dfrac {2^{64} - 1} {2 - 1} = 2^{64} - 1$

Let it be assumed that there are $2^{10} \times 3^2$ grains of wheat to the (imperial) pint.

This approximation is suggested in of $1881$.

There are $64$ (imperial) pints to the bushel.

Hence we have:

Using a conversion factor of $27.496$ bushels per cubic metre, this gives:


 * $W \approx 1 \, 137 \, 000 \, 000 \, 000 \ \mathrm m^3$

That is quite a lot of grain.