Answered

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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 21? 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.)How many grains of wheat should be placed on square 21 ?  Part 2How many total grains of wheat should be on the board after the the grains of wheat have been placed on square 21 ?Part 3What is the total weight of all the grains of wheat on the board after the grains of wheat have been placed on square 21 ?  (Round to the nearest tenth as needed.)

Sagot :

ANSWER:

Part 1: 1048576 grains

Part 2: 2097151 grains

Part 3: 299.6 lb

STEP-BY-STEP EXPLANATION:

We know from the statement that the number of wheat in each square is doubling since it starts at 1, then 2, then 4, then 8...

Therefore, we have the following sequence:

[tex]\begin{gathered} s=1,2,4,8... \\ \\ \text{ It can be written as follows:} \\ \\ s=2^0,2^1,2^2,2^3... \\ \\ \text{ Therefore, we would be left like this:} \\ \\ s=2^{n-1} \end{gathered}[/tex]

Part 1:

We have that in square 21, n = 21, therefore, we replace:

[tex]\begin{gathered} N=20^{21-1} \\ \\ N=2^{20}=1048576 \end{gathered}[/tex]

So, there will be 524288 grains of wheat in square 21

Part 2:

We calculate the sum as follows:

[tex]\begin{gathered} S=a_0\cdot\frac{r^n-1}{r-1} \\ \\ a_0=1,r=2,n=21 \\ \\ \text{ We substitute each value and calculate the sum as follows:} \\ \\ S=1\cdot\frac{2^{21}-1}{2-1} \\ \\ S=2097151 \end{gathered}[/tex]

This means that there are a total of 2097151 grains of wheat.

Part 3:

The weight is calculated by converting the number of grains of wheat into pounds, like this:

[tex]w=2097151\text{ gr}\cdot\frac{\frac{1}{7000}\text{ lb}}{1\text{ gr}}=299.6\text{ lb}[/tex]

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