Answer :
Let's analyze each summation step-by-step to determine the one that represents the total number of pennies on the chessboard:
### Summation 1:
[tex]\[ \sum_{n=1}^{64} 1 \cdot 2^{n-1} \][/tex]
This summation is representing the idea of placing one penny on the first square of a chessboard, and then doubling the number of pennies on each subsequent square. A chessboard has 64 squares. Let's calculate the total number of pennies:
[tex]\[ \begin{align*} \text{First term: } & 1 \cdot 2^{0} = 1 \\ \text{Second term: } & 1 \cdot 2^{1} = 2 \\ \text{Third term: } & 1 \cdot 2^{2} = 4 \\ \text{Fourth term: } & 1 \cdot 2^{3} = 8 \\ &\vdots \\ \text{Last term: } & 1 \cdot 2^{63} \end{align*} \][/tex]
This summation is:
[tex]\[ 1 + 2 + 4 + \cdots + 2^{63} \][/tex]
This is a geometric series with the first term [tex]\(a = 1\)[/tex] and common ratio [tex]\(r = 2\)[/tex]. The sum of the first [tex]\(n\)[/tex] terms of a geometric series can be calculated using the formula:
[tex]\[ S_n = a \frac{r^n - 1}{r - 1} \][/tex]
For [tex]\(n = 64\)[/tex]:
[tex]\[ S_{64} = 1 \cdot \frac{2^{64} - 1}{2 - 1} = 2^{64} - 1 \][/tex]
Hence, the total number of pennies on the chessboard is:
[tex]\[ 2^{64} - 1 = 18446744073709551615 \][/tex]
### Summation 2:
[tex]\[ \sum_{n=1}^{32} 2^{31} \cdot 2^{n-1} \][/tex]
Now, let's analyze the second summation. Here, [tex]\(2^{31}\)[/tex] is a constant factor multiplied by the geometric series [tex]\(2^{n-1}\)[/tex] over 32 terms.
[tex]\[ \begin{align*} \text{First term: } & 2^{31} \cdot 2^{0} = 2^{31} \\ \text{Second term: } & 2^{31} \cdot 2^{1} = 2^{32} \\ \text{Third term: } & 2^{31} \cdot 2^{2} = 2^{33} \\ \text{Fourth term: } & 2^{31} \cdot 2^{3} = 2^{34} \\ &\vdots \\ \text{Last term: } & 2^{31} \cdot 2^{31} \end{align*} \][/tex]
This sum is:
[tex]\[ 2^{31} + 2^{32} + 2^{33} + \cdots + 2^{62} \][/tex]
This is a geometric series with the first term [tex]\(a = 2^{31}\)[/tex] and common ratio [tex]\(r = 2\)[/tex]. The sum of the first [tex]\(n\)[/tex] terms of a geometric series can be calculated using the formula:
[tex]\[ S_n = a \frac{r^n - 1}{r - 1} \][/tex]
For [tex]\(n = 32\)[/tex]:
[tex]\[ S_{32} = 2^{31} \cdot \frac{2^{32} - 1}{2 - 1} = 2^{31} \cdot (2^{32} - 1) \][/tex]
Hence:
[tex]\[ 2^{31} \cdot (2^{32} - 1) = 9223372034707292160 \][/tex]
### Summation 3:
[tex]\[ \sum_{n=1}^{32} 2^{32} \cdot 2^{n-1} \][/tex]
Finally, let's analyze the third summation. Here, [tex]\(2^{32}\)[/tex] is a constant factor multiplied by the geometric series [tex]\(2^{n-1}\)[/tex] over 32 terms.
[tex]\[ \begin{align*} \text{First term: } & 2^{32} \cdot 2^{0} = 2^{32} \\ \text{Second term: } & 2^{32} \cdot 2^{1} = 2^{33} \\ \text{Third term: } & 2^{32} \cdot 2^{2} = 2^{34} \\ \text{Fourth term: } & 2^{32} \cdot 2^{3} = 2^{35} \\ &\vdots \\ \text{Last term: } & 2^{32} \cdot 2^{31} \end{align*} \][/tex]
This sum is:
[tex]\[ 2^{32} + 2^{33} + 2^{34} + \cdots + 2^{63} \][/tex]
This is a geometric series with the first term [tex]\(a = 2^{32}\)[/tex] and common ratio [tex]\(r = 2\)[/tex]. The sum of the first [tex]\(n\)[/tex] terms of a geometric series can be calculated using the formula:
[tex]\[ S_n = a \frac{r^n - 1}{r - 1} \][/tex]
For [tex]\(n = 32\)[/tex]:
[tex]\[ S_{32} = 2^{32} \cdot \frac{2^{32} - 1}{2 - 1} = 2^{32} \cdot (2^{32} - 1) \][/tex]
Hence:
[tex]\[ 2^{32} \cdot (2^{32} - 1) = 18446744069414584320 \][/tex]
### Conclusion:
The summation that represents the total number of pennies on the chessboard is:
[tex]\[ \sum_{n=1}^{64} 1 \cdot 2^{n-1} \][/tex]
which equals [tex]\(18446744073709551615\)[/tex] pennies.
### Summation 1:
[tex]\[ \sum_{n=1}^{64} 1 \cdot 2^{n-1} \][/tex]
This summation is representing the idea of placing one penny on the first square of a chessboard, and then doubling the number of pennies on each subsequent square. A chessboard has 64 squares. Let's calculate the total number of pennies:
[tex]\[ \begin{align*} \text{First term: } & 1 \cdot 2^{0} = 1 \\ \text{Second term: } & 1 \cdot 2^{1} = 2 \\ \text{Third term: } & 1 \cdot 2^{2} = 4 \\ \text{Fourth term: } & 1 \cdot 2^{3} = 8 \\ &\vdots \\ \text{Last term: } & 1 \cdot 2^{63} \end{align*} \][/tex]
This summation is:
[tex]\[ 1 + 2 + 4 + \cdots + 2^{63} \][/tex]
This is a geometric series with the first term [tex]\(a = 1\)[/tex] and common ratio [tex]\(r = 2\)[/tex]. The sum of the first [tex]\(n\)[/tex] terms of a geometric series can be calculated using the formula:
[tex]\[ S_n = a \frac{r^n - 1}{r - 1} \][/tex]
For [tex]\(n = 64\)[/tex]:
[tex]\[ S_{64} = 1 \cdot \frac{2^{64} - 1}{2 - 1} = 2^{64} - 1 \][/tex]
Hence, the total number of pennies on the chessboard is:
[tex]\[ 2^{64} - 1 = 18446744073709551615 \][/tex]
### Summation 2:
[tex]\[ \sum_{n=1}^{32} 2^{31} \cdot 2^{n-1} \][/tex]
Now, let's analyze the second summation. Here, [tex]\(2^{31}\)[/tex] is a constant factor multiplied by the geometric series [tex]\(2^{n-1}\)[/tex] over 32 terms.
[tex]\[ \begin{align*} \text{First term: } & 2^{31} \cdot 2^{0} = 2^{31} \\ \text{Second term: } & 2^{31} \cdot 2^{1} = 2^{32} \\ \text{Third term: } & 2^{31} \cdot 2^{2} = 2^{33} \\ \text{Fourth term: } & 2^{31} \cdot 2^{3} = 2^{34} \\ &\vdots \\ \text{Last term: } & 2^{31} \cdot 2^{31} \end{align*} \][/tex]
This sum is:
[tex]\[ 2^{31} + 2^{32} + 2^{33} + \cdots + 2^{62} \][/tex]
This is a geometric series with the first term [tex]\(a = 2^{31}\)[/tex] and common ratio [tex]\(r = 2\)[/tex]. The sum of the first [tex]\(n\)[/tex] terms of a geometric series can be calculated using the formula:
[tex]\[ S_n = a \frac{r^n - 1}{r - 1} \][/tex]
For [tex]\(n = 32\)[/tex]:
[tex]\[ S_{32} = 2^{31} \cdot \frac{2^{32} - 1}{2 - 1} = 2^{31} \cdot (2^{32} - 1) \][/tex]
Hence:
[tex]\[ 2^{31} \cdot (2^{32} - 1) = 9223372034707292160 \][/tex]
### Summation 3:
[tex]\[ \sum_{n=1}^{32} 2^{32} \cdot 2^{n-1} \][/tex]
Finally, let's analyze the third summation. Here, [tex]\(2^{32}\)[/tex] is a constant factor multiplied by the geometric series [tex]\(2^{n-1}\)[/tex] over 32 terms.
[tex]\[ \begin{align*} \text{First term: } & 2^{32} \cdot 2^{0} = 2^{32} \\ \text{Second term: } & 2^{32} \cdot 2^{1} = 2^{33} \\ \text{Third term: } & 2^{32} \cdot 2^{2} = 2^{34} \\ \text{Fourth term: } & 2^{32} \cdot 2^{3} = 2^{35} \\ &\vdots \\ \text{Last term: } & 2^{32} \cdot 2^{31} \end{align*} \][/tex]
This sum is:
[tex]\[ 2^{32} + 2^{33} + 2^{34} + \cdots + 2^{63} \][/tex]
This is a geometric series with the first term [tex]\(a = 2^{32}\)[/tex] and common ratio [tex]\(r = 2\)[/tex]. The sum of the first [tex]\(n\)[/tex] terms of a geometric series can be calculated using the formula:
[tex]\[ S_n = a \frac{r^n - 1}{r - 1} \][/tex]
For [tex]\(n = 32\)[/tex]:
[tex]\[ S_{32} = 2^{32} \cdot \frac{2^{32} - 1}{2 - 1} = 2^{32} \cdot (2^{32} - 1) \][/tex]
Hence:
[tex]\[ 2^{32} \cdot (2^{32} - 1) = 18446744069414584320 \][/tex]
### Conclusion:
The summation that represents the total number of pennies on the chessboard is:
[tex]\[ \sum_{n=1}^{64} 1 \cdot 2^{n-1} \][/tex]
which equals [tex]\(18446744073709551615\)[/tex] pennies.
