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Papessa [141]
3 years ago
7

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.)
Mathematics
1 answer:
Sonja [21]3 years ago
6 0

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

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