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, what is the total weight in tons of all the wheat that will be placed on the first 45 squares?
(Assume that each grain of wheat weighs 1/7000 pound. Remember that 1 ton equals 2000 lbs.)
The total weight of the wheat that will be placed on the first 45 squares is
nothing tons.

King rewarded the inventor of chess by giving grain of wheat on every square of the chess board. He places one grain of wheat on first square, 2 on second, 4 on 3rd, 8 on 4th and so on.

In fact he places grains of wheat in a progression S = 1, 2, 4, 8, 16.........n terms where value of n is 64.

We can rewrite the progression as S = 2°, 2, 2², 2³,.........n terms (n = 64)

In a geometric progression we know the nth term = [tex]a(r^{n-1})[/tex]

where a = first term of the series, r = common ratio, n = number of terms

Now we have to calculate the weight of all wheat placed on 45th square.

So we will put the values in the given formula 45th term =

[tex]1\times (2^{45-1})[/tex] = [tex]2^{44}[/tex]=[tex]1.76\times 10^{13}[/tex]

Then the weight of wheat will be = [tex]2.51\times 10^{9}pounds[/tex]=[tex]1.26\times 10^{6}tons[/tex]