Question

There are 8 rows and 8 columns, or 64 squares on a chessboard. suppose you place 1 penny on row 1 column a, 2 pennies on row 1 column b, 4 pennies on row 1 column c, and so on …

Answer 1

Though you do not finish the question, I imagine the problem is to either find the number of pennies on a particular square, or the number of pennies on the last square or the sum of the pennies once the board is full.

No matter which we are dealing with a sequence. The number of coins doubles each time. So we have: 1, 2, 4, 8, 16, 32, 64, ...

This is called a geometric sequence. To "move" from one term (number) to the next we multiply by two. The number we multiply by (2) is called the common ratio (call it r).

If we call the first term $a_{1}$, the next $a_{2}$ and so on you can write the terms like this:

$a_{1}, a_{2}r, a_{2}r ^{2} , a_{2}r ^{3} , ...$

So the nth term is given by $a_{1} r ^{n-1}$

If you are asked for the 64th term (the number of pennies on the last square) this is given by $(1) (2) ^{64-1}=2 ^{63}$ which you might have to leave like that as your calculator may not be able to handle it.

If you are asked for the number of pennies on some other square you use that number (5 for the 5th square) for n instead.

The sum of the terms of a geometric sequence is given by $\frac{ a_{1}( r^{n}-1 )}{r-1}$ so if you are being asked for the sum of the pennies on the board (64 square so you use 64 for n) you get $\frac{ 1( 2^{64}-1 )}{2-1}= 2^{64}-1$

If you are asked for the sum of the first 5 squares then use 5 for n instead and so on.

Answer 2

Answer:

The number of pennies doubles each time, so the count grows exponentially.

The number of pennies on Rows 5-8 will be much greater than on Rows 1-4.