Question

In an old fable, a commoner who had saved the king's life was told he could ask the king for any just reward. Being a shrewd man, the commoner said, "A simple wish, sire. Place one grain of wheat on the first square of a chessboard, two grains on the second square, four grains on the third square, continuing until you have filled the board. This is all I seek." Compute the total number of grains needed to do this to see why the request, seemingly simple, could not be granted. (A chessboard consists of \(8 \times 8=64\) squares.

Short answer

The total number of grains is 18,446,744,073,709,551,615.

Step-by-step solution

Understand the Chessboard and Grain Distribution

The chessboard has 64 squares. On the first square, there is 1 grain. On the second square, there are 2 grains. On the third square, there are 4 grains, and so on.

Recognize the Pattern

The number of grains on each square follows a geometric progression where each term is twice the previous one. The number of grains on the nth square is given by: \( a_n = 2^{(n-1)} \).

Summing the Series

To find the total number of grains on the chessboard, sum the geometric series: \( S = 2^0 + 2^1 + 2^2 + \dots + 2^{(64-1)} \).

Use the Formula for Sum of a Geometric Series

The sum of the first \( n \) terms of a geometric series is given by the formula: \( S_n = \frac{a(r^n-1)}{r-1} \), where \( a = 1 \), \( r = 2 \), and \( n = 64 \).

Calculate the Total Number of Grains

Plug the values into the formula: \( S_{64} = \frac{1(2^{64}-1)}{2-1} = 2^{64} - 1 \). This equals 18,446,744,073,709,551,615 grains.

Key concepts

Exponential Growth

Exponential growth occurs when a quantity increases by a consistent multiplicative rate over equal increments of time or position. This is seen in the fable where the number of wheat grains doubles with each successive square on the chessboard. Starting with just one grain of wheat on the first square, the quantity doubles on each next square: 2 grains on the second square, 4 grains on the third, 8 grains on the fourth, and so on. This rapid increase is a hallmark of exponential growth.

In mathematical terms, we express this growth using exponents. If the amount on the first square is given by 1 grain, the amount on the nth square is given by the formula: \(2^{(n-1)}\). This exponential term reflects the continuous doubling event at each step.

Exponential growth can quickly lead to huge numbers, much larger than simple linear growth would. Understanding exponential growth not only helps in solving problems like the fable's but also in real-world contexts such as population growth, finance, and technology advancements.

Geometric Progression

A geometric progression (or geometric sequence) is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In the wheat and chessboard problem, the common ratio is 2, meaning each term is twice the previous one.

Here's the sequence of the first few grains on the chessboard:

• 1 grain (first square)
• 2 grains (second square)
• 4 grains (third square)
• 8 grains (fourth square)

And it continues this way until the 64th square. This sequence is a classic example of a geometric progression.

The general formula for the nth term in a geometric progression can be written as \(a_n = a \times r^{(n-1)}\), where:

• \(a\) is the first term (1 grain of wheat)
• \(r\) is the common ratio (2)
• \(n\) is the term number

Understanding geometric progressions is crucial, as they appear in many areas, from financial calculations to physics and computer science.

Sum of a Series

The sum of a series refers to the addition of all terms in a sequence. For a geometric series like the one in the fable, there is a specific formula to calculate the sum of its terms. The fable requires us to find the total grains on all 64 squares of the chessboard. The series we sum looks like this: \(2^0 + 2^1 + 2^2 + \text{...} + 2^{63}\).

We use the formula to find the sum of the first \(n\) terms of a geometric series, which is: \( S_n = \frac{a(r^n - 1)}{r - 1} \).

For the problem at hand:

• \(a\) (the first term) is 1
• \(r\) (the common ratio) is 2
• \(n\) (the number of terms) is 64

Plugging in these values, we get: \( S_{64} = \frac{1(2^{64} - 1)}{2 - 1} = 2^{64} - 1 \). Calculating this yields 18,446,744,073,709,551,615 grains of wheat. This large number shows why the commoner's seemingly modest wish could not realistically be granted.

This principle of series summation is useful in various mathematical and practical applications, such as calculating loans, investments, and understanding natural phenomena.