Question

A King in ancient times agreed to reward the inventor of chess with one grain of wheat on the first of the 64 squares of a chess board. On the second square the King would place two grains of​ wheat, on the third​ square, four grains of​ wheat, and on the fourth square eight grains of wheat. If the amount of wheat is doubled in this way on each of the remaining​ squares, what is the total weight in tons of all the wheat that will be placed on the first 60 ​squares? (Assume that each grain of wheat weighs​ 1/7000 pound. Remember that 1 tonequals2000 ​lbs.) The total weight of the wheat that will be placed on the first 60 squares is ? tons. ​(Use scientific notation.

Answer 1

Introduction:

From the given information, observe that, number of grains of wheat placed on the r^th square (r = 1, 2, 3, …, 60) of the chess board is 2^{r – 1}. This can be verified as follows:

For r = 1: 2^1–1 = 2⁰ = 1.

For r = 2: 2^2–1 = 2¹ = 2.

For r = 3: 2^3–1 = 2² = 4.

For r = 4: 2^4–1 = 2³ = 8, and so on.

Proceeding in this manner, it can be understood that the last or the 60^th box will be: 2^60–1 = 2⁵⁹.

Note that this matches with the given information.

Calculation:

First, it is necessary to calculate the total number of wheat grains placed on the first 60 squares (say, S). This can be found by using the geometric series for a finite number of cases, as follows:

Thus, the total weight of the wheat that will be placed on the first 60 squares is 8.235 × 10¹⁰ ton.