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Sagot :
b. Generalize the Results in the Table
To generalize the results, we start with understanding the geometric series at stage [tex]\( n \)[/tex].
At stage [tex]\( n \)[/tex], the second column gives us the terms of the geometric series:
[tex]\[ \frac{3}{4} + \frac{3}{16} + \cdots + 3 \left(\frac{1}{4}\right)^n \][/tex]
This series can be expressed as:
[tex]\[ a + ar + ar^2 + \cdots + ar^n \][/tex]
Where:
- [tex]\( a = \frac{3}{4} \)[/tex] (the first term)
- [tex]\( r = \frac{1}{4} \)[/tex] (the common ratio)
The sum of the first [tex]\( n \)[/tex] terms of a finite geometric series is given by the formula:
[tex]\[ S_n = a \frac{1 - r^{n+1}}{1 - r} \][/tex]
By substituting [tex]\( a = \frac{3}{4} \)[/tex] and [tex]\( r = \frac{1}{4} \)[/tex] into the formula, we get the specific formula that the third column gives you:
[tex]\[ S_n = \frac{3}{4} \frac{1 - \left(\frac{1}{4}\right)^{n+1}}{1 - \frac{1}{4}} \][/tex]
Simplifying this, we obtain:
[tex]\[ S_n = \frac{3}{4} \frac{1 - \left(\frac{1}{4}\right)^{n+1}}{\frac{3}{4}} \][/tex]
This simplifies further to:
[tex]\[ S_n = 3 \left(1 - \left(\frac{1}{4}\right)^{n+1}\right) \][/tex]
So the formula given in the third column is:
[tex]\[ S_n = 3 \left(1 - \left(\frac{1}{4}\right)^{n+1}\right) \][/tex]
c. Show that the General Formula for the Sum of a Finite Geometric Series Agrees with the Specific Formula
To show that the general formula for the sum of a finite geometric series [tex]\( S_n = a \frac{1 - r^{n+1}}{1 - r} \)[/tex] agrees with the specific formula [tex]\( S_n = 3 \left(1 - (\frac{1}{4})^{n+1}\right) \)[/tex], we need to compare these formulas for specific terms.
For the general formula:
[tex]\[ S_n = a \frac{1 - r^{n+1}}{1 - r} \][/tex]
Where [tex]\( a = \frac{3}{4} \)[/tex] and [tex]\( r = \frac{1}{4} \)[/tex], we substitute these values:
[tex]\[ S_n = \frac{3}{4} \frac{1 - (\frac{1}{4})^{n+1}}{1 - \frac{1}{4}} \][/tex]
[tex]\[ S_n = \frac{3}{4} \frac{1 - (\frac{1}{4})^{n+1}}{\frac{3}{4}} \][/tex]
[tex]\[ S_n = 3 \left(1 - (\frac{1}{4})^{n+1}\right) \][/tex]
Therefore, the general formula
[tex]\[ S_n = a \frac{1 - r^{n+1}}{1 - r} \][/tex]
for [tex]\( a = \frac{3}{4} \)[/tex] and [tex]\( r = \frac{1}{4} \)[/tex] is indeed equal to the specific formula:
[tex]\[ S_n = 3(1 - (\frac{1}{4})^{n+1}) \][/tex]
Hence, the general formula for the sum of a finite geometric series perfectly agrees with the specific formula derived from part (b).
To generalize the results, we start with understanding the geometric series at stage [tex]\( n \)[/tex].
At stage [tex]\( n \)[/tex], the second column gives us the terms of the geometric series:
[tex]\[ \frac{3}{4} + \frac{3}{16} + \cdots + 3 \left(\frac{1}{4}\right)^n \][/tex]
This series can be expressed as:
[tex]\[ a + ar + ar^2 + \cdots + ar^n \][/tex]
Where:
- [tex]\( a = \frac{3}{4} \)[/tex] (the first term)
- [tex]\( r = \frac{1}{4} \)[/tex] (the common ratio)
The sum of the first [tex]\( n \)[/tex] terms of a finite geometric series is given by the formula:
[tex]\[ S_n = a \frac{1 - r^{n+1}}{1 - r} \][/tex]
By substituting [tex]\( a = \frac{3}{4} \)[/tex] and [tex]\( r = \frac{1}{4} \)[/tex] into the formula, we get the specific formula that the third column gives you:
[tex]\[ S_n = \frac{3}{4} \frac{1 - \left(\frac{1}{4}\right)^{n+1}}{1 - \frac{1}{4}} \][/tex]
Simplifying this, we obtain:
[tex]\[ S_n = \frac{3}{4} \frac{1 - \left(\frac{1}{4}\right)^{n+1}}{\frac{3}{4}} \][/tex]
This simplifies further to:
[tex]\[ S_n = 3 \left(1 - \left(\frac{1}{4}\right)^{n+1}\right) \][/tex]
So the formula given in the third column is:
[tex]\[ S_n = 3 \left(1 - \left(\frac{1}{4}\right)^{n+1}\right) \][/tex]
c. Show that the General Formula for the Sum of a Finite Geometric Series Agrees with the Specific Formula
To show that the general formula for the sum of a finite geometric series [tex]\( S_n = a \frac{1 - r^{n+1}}{1 - r} \)[/tex] agrees with the specific formula [tex]\( S_n = 3 \left(1 - (\frac{1}{4})^{n+1}\right) \)[/tex], we need to compare these formulas for specific terms.
For the general formula:
[tex]\[ S_n = a \frac{1 - r^{n+1}}{1 - r} \][/tex]
Where [tex]\( a = \frac{3}{4} \)[/tex] and [tex]\( r = \frac{1}{4} \)[/tex], we substitute these values:
[tex]\[ S_n = \frac{3}{4} \frac{1 - (\frac{1}{4})^{n+1}}{1 - \frac{1}{4}} \][/tex]
[tex]\[ S_n = \frac{3}{4} \frac{1 - (\frac{1}{4})^{n+1}}{\frac{3}{4}} \][/tex]
[tex]\[ S_n = 3 \left(1 - (\frac{1}{4})^{n+1}\right) \][/tex]
Therefore, the general formula
[tex]\[ S_n = a \frac{1 - r^{n+1}}{1 - r} \][/tex]
for [tex]\( a = \frac{3}{4} \)[/tex] and [tex]\( r = \frac{1}{4} \)[/tex] is indeed equal to the specific formula:
[tex]\[ S_n = 3(1 - (\frac{1}{4})^{n+1}) \][/tex]
Hence, the general formula for the sum of a finite geometric series perfectly agrees with the specific formula derived from part (b).
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