In a well known story the inventor of the game of chess was asked by his well pleased King what reward he desired. "Oh, not much, your majesty", the inventor responded, "just place a grain of rice on the first square of the board, 2 on the next, 4 on the next, and so on, twice as many on each square as on the preceding one. I will give this rice to the poor." (For the uninitiated, a chess board has 64 squares.) The king thought this a modest request indeed and ordered the rice to be delivered.
Let f(n) denote the number of rice grains placed on the first n squares of the board. So clearly, f(1)=1, f(2) = 1+2 = 3, f(3)= 1+ 2 + 4 =7, and so on. How does it go on? Compute the next two values of f(n): f(4) =_ ~ f(5) =_ ~
Ponder the structure of this summation and then enter an algebraic expression that defines
f(n) =____ as a function of n. Supposing that there are 25,000 grains of rice in a pound, 2000 pounds in a ton, and 6 billion people on earth, the inventor's reward would work out to approximately 3(10)^17 tons of rice for every person on the planet. Clearly, all the rice in the kingdom would not be enough to begin to fill that request. The story has a sad ending: feeling duped, the king caused the inventor of chess to be beheaded.

The amount of rice is always doubling the previous one, in other words, for the squares 1, 2, 3, and 4, for example, there will be respectively, 1, 2, 4, and 8 grains of rice. Continuing with the same idea in order to find the number of rice grains placed on the first 4 squares, we have: 1+2+4+8=15 and it is the accumulated addition to the 4th square. The last explanation can be shown in the following table writing the information for the first 6 squares.

Number of the square Amount of grains, A(n) Accumulated addition, f(n)

n

1 1 1

2 2 3

3 4 7

4 8 15

5 16 31

6 32 63

On the Amount of grains column, it can be seen a pattern. Those numbers are power of 2, because [tex]1=2^{0} , 2=2^{1}, 4=2^{2}, 8=2^{3}[/tex], and so on. As you can see, each exponent in the power is one less than the number of the square. For the 3rd square, the exponent is 2, for the 4th square, the exponent is 3. Following the pattern, for the nth-square, the exponent will be n-1. The function for A(n) is [tex]A(n)=2^{(n-1)}[/tex]

Let´s prove it. For n=7, [tex]A(7)=2^{(7-1)}=2^{6}=64[/tex]

And 64 is the double of 32 grains for the 6th square.

Now, observing the Accumulated Addition, f(n), on the third column, each number is the previous number of a power of 2. [tex]3=2^{2}-1, 7=2^{3} -1, 15=2^{4}-1[/tex], and so on. If you see the respective exponent, it coincides with the corresponding number of the square. Generalizing, the function f(n) is [tex]f(n)=2^{n}-1[/tex]

Evaluating f(4) and f(5) we get:

[tex]f(4)=2^{4}-1=16-1=15[/tex] grains.

[tex]f(5)=2^{5}-1=32-1=31[/tex] grains.

And this is all.