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

Complete 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 …

Determine the number of pennies in Row 1

Determine the number of pennies on the entire chessboard?

Answer:

255 in the first row

18,446,744,073,709,551,615 in the entire board

Step-by-step explanation:

Given

$Rows = 8$

$Columns = 8$

Solving (a): Number of pennies in first row

The question is an illustration of geometric sequence which follows

$1,2,4....$

Where

$a =1$ --- The first term

Calculate the common ratio, r

$r = \frac{T_2}{T_1} = \frac{4}{2} = 2$

The number of pennies in the first row will be calculated using sum of n terms of a GP.

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

Since, the first row has 8 columns, then

$n = 8$

Substitute 8 for n, 2 for r and 1 for a in $S_n = \frac{a(r^n - 1)}{r - 1}$

$S_8 = \frac{1 * (2^8 - 1)}{2 - 1}$

$S_8 = \frac{1 * (256 - 1)}{1}$

$S_8 = \frac{1 * 255}{1}$

$S_8 = 255$

Solving (b): The entire board has 64 cells.

So:

$n = 64$

Substitute 64 for n, 2 for r and 1 for a in $S_n = \frac{a(r^n - 1)}{r - 1}$

$S_{64} = \frac{1 * (2^{64} - 1)}{2 -1}$

$S_{64} = \frac{(2^{64} - 1)}{1}$

$S_{64} = \frac{(18,446,744,073,709,551,616 - 1)}{1}$

$S_{64} = \frac{18,446,744,073,709,551,615}{1}$

$S_{64} = 18,446,744,073,709,551,615$