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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

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Mamasu
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. 

MrRoyal

The pennies on the chessboard is an illustration of a geometric sequence.

  • There are 255 pennies on the first row
  • There are 4294967295 pennies on the first to four rows
  • There are [tex]\mathbf{1.8446744 \times 10^{19}}[/tex] pennies on the entire board

(a) Pennies in row 1

The given parameters are:

[tex]\mathbf{a = 1}[/tex] --- the first term

[tex]\mathbf{r = 2}[/tex] --- the common ratio

There are 8 squares on the first row.

So, we make use of the following sum of n terms of a geometric sequence

[tex]\mathbf{S_n = \frac{a(r^n- 1)}{r - 1}}[/tex]

This gives

[tex]\mathbf{S_8 = \frac{1 \times (2^8- 1)}{2 - 1}}[/tex]

[tex]\mathbf{S_8 = \frac{255}{1}}[/tex]

[tex]\mathbf{S_8 = 255}[/tex]

Hence, there are 255 pennies on the first row

(b) Pennies in row 1 - 4

There are 32 squares on the first to four rows.

So, we make use of the following sum of n terms of a geometric sequence

[tex]\mathbf{S_n = \frac{a(r^n- 1)}{r - 1}}[/tex]

This gives

[tex]\mathbf{S_{32} = \frac{1 \times (2^{32}- 1)}{2 - 1}}[/tex]

[tex]\mathbf{S_{32} = 4294967295}[/tex]

Hence, there are 4294967295 pennies on the first to four rows

(c) Pennies on the board

There are 64 squares on the board

So, we make use of the following sum of n terms of a geometric sequence

[tex]\mathbf{S_n = \frac{a(r^n- 1)}{r - 1}}[/tex]

This gives

[tex]\mathbf{S_{64} = \frac{1 \times (2^{64}- 1)}{2 - 1}}[/tex]

[tex]\mathbf{S_{64} = 1.8446744 \times 10^{19}}[/tex]

Hence, there are [tex]\mathbf{1.8446744 \times 10^{19}}[/tex] pennies on the entire board

Read more about geometric sequence at:

https://brainly.com/question/18109692