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, 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.

Answer 1

Answer:

Total weight of the wheat placed on 45th square will be = $1.26\times 10^{6}tons$

Step-by-step explanation:

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 = $a(r^{n-1})$

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 =

$1\times (2^{45-1})$ = $2^{44}$=$1.76\times 10^{13}$

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