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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.)

Respuesta :

calculista

Answer:

Part 1) On square 18 should be place [tex]131,072\ grains[/tex]

Part 2) The total number of grains of wheat is [tex]18,446,744,073,709,551,615\ grains[/tex]

Part 3) The total weight is [tex]W=2,635,249,153,387,079\ pounds[/tex]

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

[tex]an=a1(r^{n-1} )[/tex]

For n=18

we have

a1=1

r=2

substitute

[tex]a18=(1)(2^{18-1})[/tex]

[tex]a18=(2^{17})[/tex]

[tex]a18=131,072\ grains[/tex]

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

[tex]S=a1\frac{1-r^{n}}{1-r}[/tex]

we have

a1=1

r=2

n=64

substitute

[tex]S=(1)\frac{1-2^{64}}{1-2}[/tex]

[tex]S=\frac{1-2^{64}}{-1}[/tex]

[tex]S=18,446,744,073,709,551,615\ grains[/tex]

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)

[tex]W=2,635,249,153,387,078.8\ pounds[/tex]

Round to the nearest pound

[tex]W=2,635,249,153,387,079\ pounds[/tex]

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