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 16​? 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

Solution:

On the first square the king would place one grain of wheat, on the second square two grains of wheat, on the third square four grains of wheat and on the fourth square eight grains of wheat.

We can see that the series is like 1, 2, 4, 8,....

i.e. 2⁰, 2¹, 2², 2³,...

1^st term is 2⁰ , 2^nd term is 2¹

The n^th term will be 2^n-1

So, the 16^th term = 2^16-1 = 2¹⁵ = 32,768

On the square 16 the number of grains of wheat should be 2¹⁵ or 32,768

There is total 64 square on the chess board.

The total number of grains of wheat on the board = 1 + 12¹ + 12² + 12³ +.......+ 12^(64-1)

It is a geometric progression(GP) series

Formula for the sum of first 'n' number of terms in a geometric progression(a, ar, ar²,..., ar^n-1) is-

where, a = the first term

r = the common ratio

The sum of first 64 terms in GP is-

here, a = 1 and r = 2

The total number of grains of wheat on the board is 1.844710¹⁹

Weight of each grain of wheat = 1/7000 pound

The total weight of the grains of wheat in pounds = (1.844710¹⁹)(1/7000) = 2.63510¹⁵ pounds