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Sagot :
Sure, let's find the sum of the terms of the sequence given by [tex]\(\log(x), \log(x^2), \log(x^3), \dots, \log(x^{15})\)[/tex].
This sequence is in the form [tex]\(\log(x^k)\)[/tex] where [tex]\(k\)[/tex] ranges from 1 to 15.
### Step-by-Step Solution:
1. Identify the terms:
The terms of the sequence are:
[tex]\[ \log(x), \log(x^2), \log(x^3), \ldots, \log(x^{15}) \][/tex]
2. Use properties of logarithms:
Recall the property of logarithms: [tex]\(\log(a^b) = b \cdot \log(a)\)[/tex]. Using this property, we can rewrite each term:
[tex]\[ \log(x^k) = k \cdot \log(x) \][/tex]
Therefore, the sequence becomes:
[tex]\[ 1 \cdot \log(x), 2 \cdot \log(x), 3 \cdot \log(x), \ldots, 15 \cdot \log(x) \][/tex]
3. Extract [tex]\(\log(x)\)[/tex] as a common factor:
Since [tex]\(\log(x)\)[/tex] is common in all terms, factor it out from the sum:
[tex]\[ \sum_{k=1}^{15} k \cdot \log(x) = \log(x) \cdot \sum_{k=1}^{15} k \][/tex]
4. Sum of the first 15 natural numbers:
The sum of the first [tex]\(n\)[/tex] natural numbers is given by the formula:
[tex]\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \][/tex]
Substituting [tex]\(n = 15\)[/tex]:
[tex]\[ \sum_{k=1}^{15} k = \frac{15 \cdot 16}{2} = 120 \][/tex]
5. Combine results:
Now we have:
[tex]\[ \sum_{k=1}^{15} k \cdot \log(x) = \log(x) \cdot 120 \][/tex]
6. Conclusion:
Thus, the sum of the terms of the given sequence can be written as:
[tex]\[ 120 \log(x) \][/tex]
To match the provided format, if [tex]\(n\)[/tex] is denoted as the number of terms, in this case, [tex]\(n = 15\)[/tex], we can generally state the result as:
[tex]\[ \frac{n(n + 1)}{2} \log(x) \][/tex]
For [tex]\(n = 15\)[/tex]:
[tex]\[ \frac{15 \cdot 16}{2} \log(x) = 120 \log(x) \][/tex]
So, the sum of the terms of [tex]\(\log(x), \log(x^2), \cdot \cdot \cdot, \log(x^{15})\)[/tex] is [tex]\(120 \log(x)\)[/tex].
This sequence is in the form [tex]\(\log(x^k)\)[/tex] where [tex]\(k\)[/tex] ranges from 1 to 15.
### Step-by-Step Solution:
1. Identify the terms:
The terms of the sequence are:
[tex]\[ \log(x), \log(x^2), \log(x^3), \ldots, \log(x^{15}) \][/tex]
2. Use properties of logarithms:
Recall the property of logarithms: [tex]\(\log(a^b) = b \cdot \log(a)\)[/tex]. Using this property, we can rewrite each term:
[tex]\[ \log(x^k) = k \cdot \log(x) \][/tex]
Therefore, the sequence becomes:
[tex]\[ 1 \cdot \log(x), 2 \cdot \log(x), 3 \cdot \log(x), \ldots, 15 \cdot \log(x) \][/tex]
3. Extract [tex]\(\log(x)\)[/tex] as a common factor:
Since [tex]\(\log(x)\)[/tex] is common in all terms, factor it out from the sum:
[tex]\[ \sum_{k=1}^{15} k \cdot \log(x) = \log(x) \cdot \sum_{k=1}^{15} k \][/tex]
4. Sum of the first 15 natural numbers:
The sum of the first [tex]\(n\)[/tex] natural numbers is given by the formula:
[tex]\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \][/tex]
Substituting [tex]\(n = 15\)[/tex]:
[tex]\[ \sum_{k=1}^{15} k = \frac{15 \cdot 16}{2} = 120 \][/tex]
5. Combine results:
Now we have:
[tex]\[ \sum_{k=1}^{15} k \cdot \log(x) = \log(x) \cdot 120 \][/tex]
6. Conclusion:
Thus, the sum of the terms of the given sequence can be written as:
[tex]\[ 120 \log(x) \][/tex]
To match the provided format, if [tex]\(n\)[/tex] is denoted as the number of terms, in this case, [tex]\(n = 15\)[/tex], we can generally state the result as:
[tex]\[ \frac{n(n + 1)}{2} \log(x) \][/tex]
For [tex]\(n = 15\)[/tex]:
[tex]\[ \frac{15 \cdot 16}{2} \log(x) = 120 \log(x) \][/tex]
So, the sum of the terms of [tex]\(\log(x), \log(x^2), \cdot \cdot \cdot, \log(x^{15})\)[/tex] is [tex]\(120 \log(x)\)[/tex].
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