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Sagot :
To determine how many terms of the series [tex]\(2 + 4 + 6 + 8 + \ldots\)[/tex] must be taken so that the sum is [tex]\(100\)[/tex], we follow these steps:
1. Identify the Series:
This is an arithmetic series where the first term ([tex]\(a\)[/tex]) is [tex]\(2\)[/tex] and the common difference ([tex]\(d\)[/tex]) is also [tex]\(2\)[/tex].
2. Formulate the Sum of the First [tex]\(n\)[/tex] Terms:
The sum of the first [tex]\(n\)[/tex] terms ([tex]\(S_n\)[/tex]) of an arithmetic series can be calculated using the formula:
[tex]\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \][/tex]
Substituting the values [tex]\(a = 2\)[/tex] and [tex]\(d = 2\)[/tex], the formula becomes:
[tex]\[ S_n = \frac{n}{2} \left(2 \cdot 2 + (n-1) \cdot 2\right) = \frac{n}{2} \left(4 + 2n - 2\right) = \frac{n}{2} (2n + 2) = n (n + 1) \][/tex]
3. Set the Sum Equal to 100:
We need to find [tex]\(n\)[/tex] such that the sum [tex]\(S_n\)[/tex] equals 100:
[tex]\[ n(n + 1) = 100 \][/tex]
4. Solve the Quadratic Equation:
Rewrite the equation as a standard quadratic equation:
[tex]\[ n^2 + n - 100 = 0 \][/tex]
Solving a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] can be done using the quadratic formula:
[tex]\[ n = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \][/tex]
Here, [tex]\(a = 1\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(c = -100\)[/tex]. Substituting these into the formula gives:
[tex]\[ n = \frac{{-1 \pm \sqrt{{1^2 - 4 \cdot 1 \cdot (-100)}}}}{2 \cdot 1} = \frac{{-1 \pm \sqrt{{1 + 400}}}}{2} = \frac{{-1 \pm \sqrt{401}}}{2} \][/tex]
5. Interpret the Results:
This gives two solutions:
[tex]\[ n = \frac{{-1 + \sqrt{401}}}{2} \quad \text{and} \quad n = \frac{{-1 - \sqrt{401}}}{2} \][/tex]
Since [tex]\(n\)[/tex] represents the number of terms and must be a positive integer, we discard the negative solution:
[tex]\[ n = \frac{{-1 + \sqrt{401}}}{2} \][/tex]
Simplifying numerically, we get:
[tex]\[ n \approx \frac{{20.02498438 - 1}}{2} \approx \frac{{19.02498438}}{2} \approx 9.51249219 \][/tex]
However, since the number of terms [tex]\(n\)[/tex] must be an integer, we consider the nearest integers to check which satisfies the sum exactly. Based on checking:
- If [tex]\(n = 9\)[/tex], then the sum [tex]\(S_9\)[/tex] is:
[tex]\[ S_9 = 9(9 + 1) = 90 \quad (\text{not equal to } 100) \][/tex]
- If [tex]\(n = 10\)[/tex], then the sum [tex]\(S_{10}\)[/tex] is:
[tex]\[ S_{10} = 10(10 + 1) = 110 \quad (\text{exceeds } 100) \][/tex]
The exact scenario leads us to realize the possible error in integer consideration. Nevertheless mathematically closest solution intermediate as:
The solutions derived are:
\[
n = \frac{{-1 + \sqrt{401}}}{2} \approx 9.51249219 \text{ yet leading around 9~10}
coersively best rationalised solution nearer the accurate is inline 10 for functionnato per integeric utilization.
1. Identify the Series:
This is an arithmetic series where the first term ([tex]\(a\)[/tex]) is [tex]\(2\)[/tex] and the common difference ([tex]\(d\)[/tex]) is also [tex]\(2\)[/tex].
2. Formulate the Sum of the First [tex]\(n\)[/tex] Terms:
The sum of the first [tex]\(n\)[/tex] terms ([tex]\(S_n\)[/tex]) of an arithmetic series can be calculated using the formula:
[tex]\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \][/tex]
Substituting the values [tex]\(a = 2\)[/tex] and [tex]\(d = 2\)[/tex], the formula becomes:
[tex]\[ S_n = \frac{n}{2} \left(2 \cdot 2 + (n-1) \cdot 2\right) = \frac{n}{2} \left(4 + 2n - 2\right) = \frac{n}{2} (2n + 2) = n (n + 1) \][/tex]
3. Set the Sum Equal to 100:
We need to find [tex]\(n\)[/tex] such that the sum [tex]\(S_n\)[/tex] equals 100:
[tex]\[ n(n + 1) = 100 \][/tex]
4. Solve the Quadratic Equation:
Rewrite the equation as a standard quadratic equation:
[tex]\[ n^2 + n - 100 = 0 \][/tex]
Solving a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] can be done using the quadratic formula:
[tex]\[ n = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \][/tex]
Here, [tex]\(a = 1\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(c = -100\)[/tex]. Substituting these into the formula gives:
[tex]\[ n = \frac{{-1 \pm \sqrt{{1^2 - 4 \cdot 1 \cdot (-100)}}}}{2 \cdot 1} = \frac{{-1 \pm \sqrt{{1 + 400}}}}{2} = \frac{{-1 \pm \sqrt{401}}}{2} \][/tex]
5. Interpret the Results:
This gives two solutions:
[tex]\[ n = \frac{{-1 + \sqrt{401}}}{2} \quad \text{and} \quad n = \frac{{-1 - \sqrt{401}}}{2} \][/tex]
Since [tex]\(n\)[/tex] represents the number of terms and must be a positive integer, we discard the negative solution:
[tex]\[ n = \frac{{-1 + \sqrt{401}}}{2} \][/tex]
Simplifying numerically, we get:
[tex]\[ n \approx \frac{{20.02498438 - 1}}{2} \approx \frac{{19.02498438}}{2} \approx 9.51249219 \][/tex]
However, since the number of terms [tex]\(n\)[/tex] must be an integer, we consider the nearest integers to check which satisfies the sum exactly. Based on checking:
- If [tex]\(n = 9\)[/tex], then the sum [tex]\(S_9\)[/tex] is:
[tex]\[ S_9 = 9(9 + 1) = 90 \quad (\text{not equal to } 100) \][/tex]
- If [tex]\(n = 10\)[/tex], then the sum [tex]\(S_{10}\)[/tex] is:
[tex]\[ S_{10} = 10(10 + 1) = 110 \quad (\text{exceeds } 100) \][/tex]
The exact scenario leads us to realize the possible error in integer consideration. Nevertheless mathematically closest solution intermediate as:
The solutions derived are:
\[
n = \frac{{-1 + \sqrt{401}}}{2} \approx 9.51249219 \text{ yet leading around 9~10}
coersively best rationalised solution nearer the accurate is inline 10 for functionnato per integeric utilization.
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