Question 78547
If you look at the relationship between the number of wheat grains on the squares you'll see that the number is growing exponentially. Going from 1 to 2, the number is doubled. Continuing from 2 to 4, again the number is doubled. So to get to any term you must double the previous value. So this problem involves powers and exponents. If you wanted to get to the 10th term, you would start at the first term and double each term to get to the tenth term. To get to the 10th term, you must multiply 2 by itself 10 times. So to get to the 64th term, you must multiply 2 by itself 64 times, see the pattern? To get the pattern down officially, the sequence is {{{2^n}}} where n is any term you pick. We start off at n=0 to get the first term of 1 and we move from there. Hope a little background on this helps, so here we go.



a)To find out how many grains of wheat are on the 24th square, simply evaluate 2^23 (we go to the n-1 term since we started off at n=0).



{{{2^23=8388608}}}

If you want to verify, you can double 1 to 2, 2 to 4, etc until you get there (but it might take a while).


So the farmer would have to place 8,388,608 grains of wheat on the 24th square alone




b)
To find the sum of any geometric series, i.e. how to find 1+2+4+8+...2,147,483,648, you would use the sum of a geometric series formula.  If I have a series of n terms the sum is
{{{S=(a(1-r^(n)))/(1-r)}}} don't worry about a, we will ignore it, a=1 right now
and r is the factor to go from term to term, in this case r=2 (we're doubling everything). 
So {{{S=(1-2^24)/(1-2)}}}
{{{S=-16777215/-1}}}
{{{S=16777215}}} 

So if the checkerboard had only 24 squares, then the farmer would have to place 16,777,215 grains of wheat on the board.




c)To find how many grains of wheat he would need to fill the entire board, use the same formula but with 64 squares. 
{{{S=(1-2^64)/(1-2)}}}
{{{S=(-18446744073709551615)/(-1)}}}
{{{S=(18446744073709551615)}}}
So there are a total 18,446,744,073,709,551,615 grains of wheat needed to fill the entire board.