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 18​? 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:

Part 1) On square 18 should be place $131,072\ grains$

Part 2) The total number of grains of wheat is $18,446,744,073,709,551,615\ grains$

Part 3) The total weight is $W=2,635,249,153,387,079\ pounds$

Step-by-step explanation:

we know that

In a Geometric Sequence each term is found by multiplying the previous term by a constant

Part A) How many grains of wheat should be placed on square 18​?

In this problem we have

a1=1

a2=2

a3=4

a4=8

The common ratio (r) is equal to

a2/a1=2/1=2

a3/a2=4/2=2

a4/a3=8/4=2

so

r=2

the explicit rule for the nth term is equal to

$an=a1(r^{n-1} )$

For n=18

we have

a1=1

r=2

substitute

$a18=(1)(2^{18-1})$

$a18=(2^{17})$

$a18=131,072\ grains$

Part 2) Find the total number of grains of wheat on the board

we know that

The formula of the sum in a geometric sequence is equal to

$S=a1\frac{1-r^{n}}{1-r}$

we have

a1=1

r=2

n=64

substitute

$S=(1)\frac{1-2^{64}}{1-2}$

$S=\frac{1-2^{64}}{-1}$

$S=18,446,744,073,709,551,615\ grains$

Part 3) Find the total weight in pounds.​ (Assume that each grain of wheat weighs​ 1/7000 pound.)

To obtain the total weight multiply the total grains by (1/7,000)

$W=2,635,249,153,387,078.8\ pounds$

Round to the nearest pound

$W=2,635,249,153,387,079\ pounds$