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 51 squares? (Assume that each grain of wheat weighs 1/7000 pound. Remember that 1 ton equals 2000 lbs.)

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SaniShahbaz SaniShahbaz · Answer 1 · 2019-07-02T13:30:39+02:00

Answer:

Step-by-step explanation:

The number of grains of wheat on the n(th) square is 2^(n-1), or 2 to

the power of n-1. This is because the first square has 2^0 = 1 grain,

the second has 2^1 = 2, and the n(th) square has twice as many as the

previous. Thus the total number of grains of wheat is

S = 1 + 2 + 4 + 8 + ... + 2^63.

Since this is a geometric sequence with common ratio 2, the sum is

2^64 - 1

S = -------- = 2^64 - 1 = 18446744073709551615.

2 - 1