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Sagot :
To solve for the sums of the given geometric series and then order them from least to greatest, we follow these steps:
### Step 1: Calculate the sum of the first series
[tex]\[ \sum_{k=0}^6 27\left(\frac{1}{3}\right)^k \][/tex]
This is a geometric series with the first term [tex]\( a = 27 \)[/tex] and common ratio [tex]\( r = \frac{1}{3} \)[/tex]. The number of terms [tex]\( n = 7 \)[/tex]. The sum [tex]\( S_n \)[/tex] of a geometric series is given by:
[tex]\[ S_n = a \frac{1 - r^n}{1 - r} \][/tex]
Substituting the known values:
[tex]\[ S_7 = 27 \frac{1 - \left(\frac{1}{3}\right)^7}{1 - \frac{1}{3}} \][/tex]
Simplifying:
[tex]\[ S_7 = 27 \frac{1 - \frac{1}{2187}}{\frac{2}{3}} = 27 \cdot \frac{3}{2} \left(1 - \frac{1}{2187}\right) = 40.48148148148148 \][/tex]
### Step 2: Calculate the sum of the second series
[tex]\[ \sum_{k=0}^5 50\left(\frac{1}{8}\right)^k \][/tex]
This is a geometric series with the first term [tex]\( a = 50 \)[/tex] and common ratio [tex]\( r = \frac{1}{8} \)[/tex]. The number of terms [tex]\( n = 6 \)[/tex]. Using the sum formula:
[tex]\[ S_6 = a \frac{1 - r^n}{1 - r} \][/tex]
Substituting the values:
[tex]\[ S_6 = 50 \frac{1 - \left(\frac{1}{8}\right)^6}{1 - \frac{1}{8}} \][/tex]
Simplifying:
[tex]\[ S_6 = 50 \frac{1 - \frac{1}{262144}}{\frac{7}{8}} = 50 \cdot \frac{8}{7} \left(1 - \frac{1}{262144}\right) = 57.14263916015625 \][/tex]
### Step 3: Calculate the sum of the third series and then take the square root
[tex]\[ \sqrt{\sum_{k=0}^6 33\left(\frac{1}{2}\right)^k} \][/tex]
This is a geometric series with the first term [tex]\( a = 33 \)[/tex] and common ratio [tex]\( r = \frac{1}{2} \)[/tex]. The number of terms [tex]\( n = 7 \)[/tex]. Using the sum formula:
[tex]\[ S_7 = a \frac{1 - r^n}{1 - r} \][/tex]
Substituting the values:
[tex]\[ S_7 = 33 \frac{1 - \left(\frac{1}{2}\right)^7}{1 - \frac{1}{2}} \][/tex]
Simplifying:
[tex]\[ S_7 = 33 \frac{1 - \frac{1}{128}}{\frac{1}{2}} = 33 \cdot 2 \left(1 - \frac{1}{128}\right) = 33 \cdot 2 \cdot \frac{127}{128} = 65.015625 \][/tex]
Taking the square root:
[tex]\[ \sqrt{65.015625} = 8.092241654819757 \][/tex]
### Step 4: Order the sums from least to greatest
The calculated series sums are:
1. [tex]\( 8.092241654819757 \)[/tex]
2. [tex]\( 40.48148148148148 \)[/tex]
3. [tex]\( 57.14263916015625 \)[/tex]
Therefore, the order from least to greatest is:
[tex]\[ \sqrt{\sum_{k=0}^6 33 \left(\frac{1}{2}\right)^k} < \sum_{k=0}^6 27 \left(\frac{1}{3}\right)^k < \sum_{k=0}^5 50 \left(\frac{1}{8}\right)^k \][/tex]
### Step 1: Calculate the sum of the first series
[tex]\[ \sum_{k=0}^6 27\left(\frac{1}{3}\right)^k \][/tex]
This is a geometric series with the first term [tex]\( a = 27 \)[/tex] and common ratio [tex]\( r = \frac{1}{3} \)[/tex]. The number of terms [tex]\( n = 7 \)[/tex]. The sum [tex]\( S_n \)[/tex] of a geometric series is given by:
[tex]\[ S_n = a \frac{1 - r^n}{1 - r} \][/tex]
Substituting the known values:
[tex]\[ S_7 = 27 \frac{1 - \left(\frac{1}{3}\right)^7}{1 - \frac{1}{3}} \][/tex]
Simplifying:
[tex]\[ S_7 = 27 \frac{1 - \frac{1}{2187}}{\frac{2}{3}} = 27 \cdot \frac{3}{2} \left(1 - \frac{1}{2187}\right) = 40.48148148148148 \][/tex]
### Step 2: Calculate the sum of the second series
[tex]\[ \sum_{k=0}^5 50\left(\frac{1}{8}\right)^k \][/tex]
This is a geometric series with the first term [tex]\( a = 50 \)[/tex] and common ratio [tex]\( r = \frac{1}{8} \)[/tex]. The number of terms [tex]\( n = 6 \)[/tex]. Using the sum formula:
[tex]\[ S_6 = a \frac{1 - r^n}{1 - r} \][/tex]
Substituting the values:
[tex]\[ S_6 = 50 \frac{1 - \left(\frac{1}{8}\right)^6}{1 - \frac{1}{8}} \][/tex]
Simplifying:
[tex]\[ S_6 = 50 \frac{1 - \frac{1}{262144}}{\frac{7}{8}} = 50 \cdot \frac{8}{7} \left(1 - \frac{1}{262144}\right) = 57.14263916015625 \][/tex]
### Step 3: Calculate the sum of the third series and then take the square root
[tex]\[ \sqrt{\sum_{k=0}^6 33\left(\frac{1}{2}\right)^k} \][/tex]
This is a geometric series with the first term [tex]\( a = 33 \)[/tex] and common ratio [tex]\( r = \frac{1}{2} \)[/tex]. The number of terms [tex]\( n = 7 \)[/tex]. Using the sum formula:
[tex]\[ S_7 = a \frac{1 - r^n}{1 - r} \][/tex]
Substituting the values:
[tex]\[ S_7 = 33 \frac{1 - \left(\frac{1}{2}\right)^7}{1 - \frac{1}{2}} \][/tex]
Simplifying:
[tex]\[ S_7 = 33 \frac{1 - \frac{1}{128}}{\frac{1}{2}} = 33 \cdot 2 \left(1 - \frac{1}{128}\right) = 33 \cdot 2 \cdot \frac{127}{128} = 65.015625 \][/tex]
Taking the square root:
[tex]\[ \sqrt{65.015625} = 8.092241654819757 \][/tex]
### Step 4: Order the sums from least to greatest
The calculated series sums are:
1. [tex]\( 8.092241654819757 \)[/tex]
2. [tex]\( 40.48148148148148 \)[/tex]
3. [tex]\( 57.14263916015625 \)[/tex]
Therefore, the order from least to greatest is:
[tex]\[ \sqrt{\sum_{k=0}^6 33 \left(\frac{1}{2}\right)^k} < \sum_{k=0}^6 27 \left(\frac{1}{3}\right)^k < \sum_{k=0}^5 50 \left(\frac{1}{8}\right)^k \][/tex]
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