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Second Half of the Chessboard : ウィキペディア英語版
Wheat and chessboard problem

The wheat and chessboard problem (sometimes expressed in terms of rice instead of wheat) is a mathematical problem in the form of a word problem:
The problem may be solved using simple addition. With 64 squares on a chessboard, if the number of grains doubles on successive squares, then the sum of grains on all 64 squares is: 1 + 2 + 4 + 8... and so forth for the 64 squares. The total number of grains equals 18,446,744,073,709,551,615, much higher than what most intuitively expect.
The exercise of working through this problem may be used to explain and demonstrate exponents and the quick growth of exponential and geometric sequences. It can also be used to illustrate sigma notation.
When expressed as exponents, the geometric series is: 20 + 21 + 22
 + 23... and so forth up to 263. The base of each exponentiation, "2", expresses the doubling at each square, while the exponents represent the position of each square (0 for the first square, 1 for the second, etc.).
== Solutions ==

The simple, brute-force solution is to just manually double and add each step of the series:
: T_ = 1 + 2 + 4 + \cdots + 9,223,372,036,854,775,808 = 18,446,744,073,709,551,615
::where T_ is the total number of grains.
The series may be expressed using exponents:
: T_ = 2^ + 2^ + 2^ + \cdots + 2^
and, represented with capital-sigma notation as:
:\sum_^ 2^i.\,
It can also be solved much more easily using:
: T_ = 2^- 1. \,
A proof of which is:
: s = 2^ + 2^ + 2^ + \cdots + 2^.
Multiply each side by 2:
: 2s = 2^ + 2^ + 2^ + \cdots + 2^ + 2^.
Subtract original series from each side:
: 2s - s = - 2^ + 2^
: \therefore s = 2^- 1. \,

抄文引用元・出典: フリー百科事典『 ウィキペディア(Wikipedia)
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