Question

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.)

Answer 1

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. ​