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 20? Also find the total number of grains of wheat on the board at this time and their total weight
Here is 1 grain on square 1 , 2 on square 2, 4 on square 3 and so on. This is a geometric sequence with the nth term = a1r^(n - 1) where a1 = first term , r = common ratio.
So the 20th term (the number of grains on square 20) would be
If the answer can be left in a fraction than this is the solution: 2×5/8=5/4 5/4-1/3=11/12 if you mean the answer must be left in a whole number than.. I don't really know how to do it
