A chessboard has 64 squares. George places 1 grain of rice on the first square, 2 grains on the second square, 4 grains on the third square, 8 grains on the fourth square, and so on, until he has placed grains of rice on 10 squares.

Question

Once George has put rice on the 10th square, he has placed a total of
grains of rice on the chess board.

Answer 1

To find the total number of grains of rice George places on the first 10 squares of the chessboard, we note that the number of grains on each square doubles from the previous square. Specifically, the number of grains on the \( n \)-th square is given by:

\[ 2^{n-1} \]

where \( n \) is the square number (1 through 10).

Now we will calculate the total number of grains placed on the first 10 squares:

1. Square 1 (n=1): \( 2^{1-1} = 2^0 = 1 \)
2. Square 2 (n=2): \( 2^{2-1} = 2^1 = 2 \)
3. Square 3 (n=3): \( 2^{3-1} = 2^2 = 4 \)
4. Square 4 (n=4): \( 2^{4-1} = 2^3 = 8 \)
5. Square 5 (n=5): \( 2^{5-1} = 2^4 = 16 \)
6. Square 6 (n=6): \( 2^{6-1} = 2^5 = 32 \)
7. Square 7 (n=7): \( 2^{7-1} = 2^6 = 64 \)
8. Square 8 (n=8): \( 2^{8-1} = 2^7 = 128 \)
9. Square 9 (n=9): \( 2^{9-1} = 2^8 = 256 \)
10. Square 10 (n=10): \( 2^{10-1} = 2^9 = 512 \)

Now, we sum these values to find the total number of grains of rice:

\[ 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 \]

This series is a geometric series where the first term \( a = 1 \) and the common ratio \( r = 2 \). The sum of the first \( n \) terms of a geometric series can be calculated using the formula:

\[ S_n = a \frac{r^n - 1}{r - 1} \]

For our case (\( n = 10 \), \( a = 1 \), and \( r = 2 \)):

\[ S_{10} = 1 \frac{2^{10} - 1}{2 - 1} = 2^{10} - 1 = 1024 - 1 = 1023 \]

Thus, the total number of grains of rice George places on the first 10 squares is \( \boxed{1023} \).