Q:

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 18​? 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.)

Accepted Solution

A:
Answer:Square 15: 16,384Sum of all grains up to and including square 15: 32,767Total weight up to and including square 15: 4.681 lbStep-by-step explanation:Square 1: 1 = 2^0Square 2: 2 = 2^1Square 3: 4 = 2^2Square 4: 8 = 2^3Notice 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 = 16,384Now let's look at the sums.Square 1 sum: 1Square 2 sum: 1 + 2 = 3Square 3 sum: 3 + 4 = 7Square 4 sum: 7 + 8 = 15Notice that each sum is one grain less than 2 raised to the number of the square.Square 1 sum: 1 = 2^1 - 1Square 2 sum: 1 + 2 = 3 = 2^2 - 1Square 3 sum: 3 + 4 = 7 = 2^3 - 1Square 4 sum: 7 + 8 = 15 = 2^4 - 1Square n sum: 2^n - 1Square 15 sum: 2^15 - 1 = 32,767Weight:32,767 × 1/7000 lb = 4.681 lb