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olasank [31]
4 years ago
9

If a chessboard (8x8) were to have pennies placed on each square such that 1 penny was placed on the first square, 2 on the seco

nd, 4 on the third, and so on (doubling the number of pennies on each subsequent square), how many pennies would be on the chessboard when finished?
Mathematics
2 answers:
Lady_Fox [76]4 years ago
7 0

Answer:

18446,74,40,73.70,95,51,615

Step-by-step explanation:

There are in total 64 squares on a chess board.

I penny placed on I square, II on 2nd square, 4 on 3rd and so on.

The no of pennies on each square form a geometric sequence with each term multiplied by 2 by the previous term

Thus no of pennies placed would be

1,2,4,.....2^{n-1} ,....2^{63}

The total no of pennies placed = sum of a geometric series with I term 1 and common ratio 2

=1(\frac{2^{64}-1 }{2-1} =2^{64}-1

=18446,74,40,73.70,95,51,615

k0ka [10]4 years ago
4 0

64 according to e2020

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