MATH SOLVE

2 months ago

Q:
# 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 … Estimate: How many pennies in Row 1? pennies Estimate: How many pennies in Rows 1-4? pennies Estimate: How many pennies on the entire chessboard? pennies

Accepted Solution

A:

This is a geometric progression with a common ratio of 2.

The sum of all the pennies in the first row (8 elements) is

[tex]S_8=\frac{a_1\left(r^n-1\right)}{r-1}=\frac{1\left(2^8-1\right)}{2-1}=255[/tex]

There are 255 pennies in row 1.

In rows 1-4, there are 32 squares. So the sum is

[tex]S_{32}=\frac{a_1\left(r^n-1\right)}{r-1}=\frac{1\left(2^{32}-1\right)}{2-1}=4,294,967,295[/tex]

There are 4,294,967,295 pennies in the first 4 rows.

There are 64 squares in all. So, the sum of all pennies is

[tex]S_{64}=\frac{a_1\left(r^n-1\right)}{r-1}=\frac{1\left(2^{64}-1\right)}{2-1}=18,446,744,073,709,551,615[/tex]

There are 18,446,744,073,709,551,615 pennies on the entire chessboard.

The sum of all the pennies in the first row (8 elements) is

[tex]S_8=\frac{a_1\left(r^n-1\right)}{r-1}=\frac{1\left(2^8-1\right)}{2-1}=255[/tex]

There are 255 pennies in row 1.

In rows 1-4, there are 32 squares. So the sum is

[tex]S_{32}=\frac{a_1\left(r^n-1\right)}{r-1}=\frac{1\left(2^{32}-1\right)}{2-1}=4,294,967,295[/tex]

There are 4,294,967,295 pennies in the first 4 rows.

There are 64 squares in all. So, the sum of all pennies is

[tex]S_{64}=\frac{a_1\left(r^n-1\right)}{r-1}=\frac{1\left(2^{64}-1\right)}{2-1}=18,446,744,073,709,551,615[/tex]

There are 18,446,744,073,709,551,615 pennies on the entire chessboard.