Mathematics, 27.12.2020 07:30 damariush567

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 ...

Answers: 3

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Answer from: crispymaddix12

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?

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

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

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

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

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

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

Answer from: Quest

step-by-step explanation:

12475909

Answer from: Quest

(2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

step-by-step explanation:

simplify the following:

x/x^2 + 3 x + 2 - (x (x + 1))/(x + 2)

combine powers. x/x^2 = x^(1 - 2):

x^(1 - 2) + 3 x + 2 - (x (x + 1))/(x + 2)

1 - 2 = -1:

x^(-1) + 3 x + 2 - (x (x + 1))/(x + 2)

put each term in 1/x + 3 x + 2 - (x (x + 1))/(x + 2) over the common denominator x (x + 2): 1/x + 3 x + 2 - (x (x + 1))/(x + 2) = (x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)):

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2))

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)) = ((x + 2) + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2)):

(2 + x + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

3 x^2 (x + 2) = 3 x^3 + 6 x^2:

(2 + x + 3 x^3 + 6 x^2 + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

2 x (x + 2) = 2 x^2 + 4 x:

(2 + x + 6 x^2 + 3 x^3 + 2 x^2 + 4 x - x^2 (x + 1))/(x (x + 2))

-x^2 (x + 1) = -x^2 - x^3:

(-x^2 - x^3 + 3 x^3 + 6 x^2 + 2 x^2 + 4 x + x + 2)/(x (x + 2))

grouping like terms, 3 x^3 - x^3 + 6 x^2 + 2 x^2 - x^2 + 4 x + x + 2 = (3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2:

((3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

3 x^3 - x^3 = 2 x^3:

(2 x^3 + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

6 x^2 + 2 x^2 - x^2 = 7 x^2:

(2 x^3 + 7 x^2 + (x + 4 x) + 2)/(x (x + 2))

x + 4 x = 5 x:

answer: (2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

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