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

Answer:

Given,

Here number of grains on square :

1 2 4 8 .............

Here let us consider a geometric sequence ,

where n = 19

we know,

The nth term geometric form is given as follows

i.e.,

Tn = a*r^(n-1)

substitute values

T19 = 2^(19-1)

= 2^18

= 262144

So there are 262144 grains on the nineteenth square.

Now consider,

To give total number of grains

Sn = a(r^n - 1) / (r - 1)

substitute values

= (2^19 - 1 ) / (2 - 1)

= (524288 - 1) / 1

= 524287

So here there are 524287 total grains

In pounds , total grains = 524287 * 1/7000 [since each grain weight = 1/7000 pound]

= 74.898 pound

Total weight = 74.898