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6 September, 09:37

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 19 ? 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.)

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  1. 6 September, 13:32
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    The series is 2^ (n-1) where n=1,2,3,4, ...,62,63,64

    We can adjust the index and write it as 2^n where n=0,1,2,3,4, ...,61,62,63

    The sum of the geometric series is:

    a1 * (1 - r^n) / (1-r)

    where r is the common ratio in this case 2,

    a1 is the first term, in this case 1,

    and n is the number of term, in this case 64

    1 * (1 - 2^64) / (1-2) = 18446744073709551615

    Dividing that by 7000

    That's 2635249153387078 pounds or 1317624576693.5 tons

    grains of wheat on square 19:

    1 * (1 - 2^19) / (1-2) = 524287.
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