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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, how many grains of wheat should be placed on square ​15? Also find the total number of grains of wheat on the board at this time and their total weight in pounds.​ (Assume that each grain of wheat weighs​ 1/7000 pound.)

Sagot :

mathstudent55

Answer:

Square 15: 16,384

Sum of all grains up to and including square 15: 32,767

Total weight up to and including square 15: 4.681 lb

Step-by-step explanation:

Square 1: 1 = 2^0

Square 2: 2 = 2^1

Square 3: 4 = 2^2

Square 4: 8 = 2^3

Notice that the exponent of the 2 is 1 less than the number of the square, so for square n, the number of grains is 2^(n - 1)

Square n: 2^(n - 1)

Square 15: 2^(15 - 1) = 2^14 = 16,384

Now let's look at the sums.

Square 1 sum: 1

Square 2 sum: 1 + 2 = 3

Square 3 sum: 3 + 4 = 7

Square 4 sum: 7 + 8 = 15

Notice that each sum is one grain less than 2 raised to the number of the square.

Square 1 sum: 1 = 2^1 - 1

Square 2 sum: 1 + 2 = 3 = 2^2 - 1

Square 3 sum: 3 + 4 = 7 = 2^3 - 1

Square 4 sum: 7 + 8 = 15 = 2^4 - 1

Square n sum: 2^n - 1

Square 15 sum: 2^15 - 1 = 32,767

Weight:

32,767 × 1/7000 lb = 4.681 lb

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