How many coins are there on square 64 of a chess board starting from square 1 and doubling uo each square?

Answer 1

Square 1 has 1 coin (2^0)
Square 2 has 2 coins (2^1)
Square 3 has 4 coins (2^2)
Square 4 has 8 coins (2^3)
etc.

In general:
Square n has 2^(n-1) coins.

So:
Square 64 would have 2^63 coins.

That's about 9.22337204 × 10^18 or more than 9 quintillion coins!

(You only asked about the last square, but on the complete chessboard you would have another 9 quintillion (2^63 - 1) on squares 1 through 63 for a total of 18 quintillion (2^64 - 1) altogether... if that were your question.)

Answer 2

Square 64 has 2^63 coins on it. Here is why:

There is one coin on square one. Square 2 has double that or, 1 x 2 = 2 coins. Square three has double the number of coins on square two, or (1 x 2) x 2 = 2 x 2 = 2^2 = 4 coins. Square four has ((1 x 2) x 2) x 2 = 2 x 2 x 2 = 2^3 = 8 coins. By now you probably see the pattern. Square number n will have 2^(n-1) coins. This equation even works for square one, which has 2^(1-1) = 2^0 = 1 coin. So, square 64 has 2^(64-1) = 2^63 coins.

Answer 3

Using the mathematical formula for geometric growth, in this particular example it is doubling every instance as such.

P(x)=a(2)^n, P(x) is the final product, and a is the inial amount started with, and 2 is because it always geometrically doubles, and n is the number of times this doubling is commited.

As such this particle example would amount to P(x)=1(2)^64=

1.844674407*10^19 coins on the entire 64 squares of the chess board, so thre would be 18446744070000000000 coins on the chess board, try doing it.

On the 64th square it would be P(x)=1(2)^64 minus P(x)=1(2)^63, as such secluding the amount on the 64th square giving 9.2233720338*10^18 coins on the 64th square or 9223372033800000000 coins on the 64th square, have fun trying it out.

Answer 4

you can use the law of the sum of a geometric series:

S = { a [ (r^n) -1 ] } / (r-1)
where a is the first term, r is the ratio between two terms, and n is the number of terms.
a is 1, r=2 since you double each term, and there are 64 terms.

so S = { 1 [ (2^64) -1 ] } / (2-1) = (2^64) -1

= 1.844674407 * (10^19) = 1.845 * (10^19) coins

Answer 5

(2^64) -1 =

18.446.744.073.709.551.615

Answer 6

on the sixty fourth sq. of the chessboard there could be precisely 2 to the skill of sixty 3 = 9,223,372,036,854,775,808 grains of rice. In entire, on the finished chessboard there could be precisely 2 to the skill of sixty 4 ? a million = 18,446,744,073,709,551,615 grains of rice. to place that for the time of perspective, in simple terms the 2d a million/2 of the Chessboard might weigh 6 cases the finished Biomass of Earth.

Answer 7

1 2 4 8 16....
2^0, 2^1,2^2,2^3,2^4...2^64-1=1.844674407E19

Answer 8

None

Answer 9

Amazing!...

Answer 10

2^63 (that is a lot)

TFTP