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Sagot :
Let's tackle each of the given problems one by one:
Problem 43:
What is the [tex]\(6^{\text{th}}\)[/tex] term of a sequence whose general term is [tex]\(a_n = a_{n-1} + (n-1) a_{n-2}\)[/tex], where [tex]\(a_1 = 1\)[/tex] and [tex]\(a_2 = 3\)[/tex]?
Let's find the terms of the sequence step-by-step:
- [tex]\(a_1 = 1\)[/tex]
- [tex]\(a_2 = 3\)[/tex]
Now, using the given formula [tex]\(a_n = a_{n-1} + (n-1) a_{n-2}\)[/tex]:
- [tex]\(a_3 = a_2 + 2 \cdot a_1 = 3 + 2 \cdot 1 = 3 + 2 = 4\)[/tex]
- [tex]\(a_4 = a_3 + 3 \cdot a_2 = 4 + 3 \cdot 3 = 4 + 9 = 13\)[/tex]
- [tex]\(a_5 = a_4 + 4 \cdot a_3 = 13 + 4 \cdot 4 = 13 + 16 = 29\)[/tex]
- [tex]\(a_6 = a_5 + 5 \cdot a_4 = 29 + 5 \cdot 13 = 29 + 65 = 94\)[/tex]
Therefore, the sixth term of the sequence is:
[tex]\[ \boxed{94} \][/tex]
The correct answer is none of the given options, indicating possibly a miscalculation or error in listing the correct choices.
Problem 44:
What is the general term [tex]\(g_n\)[/tex] of the geometric sequence whose first term is [tex]\(\frac{15}{2}\)[/tex] and third term is [tex]\(\frac{135}{8}\)[/tex]?
Let's find the common ratio [tex]\(r\)[/tex]:
We know:
- [tex]\(a_1 = \frac{15}{2}\)[/tex]
- [tex]\(a_3 = \frac{135}{8}\)[/tex]
In a geometric sequence, [tex]\(a_3 = a_1 \cdot r^2\)[/tex]. Using the given values:
[tex]\[ \frac{135}{8} = \frac{15}{2} \cdot r^2 \][/tex]
Solving for [tex]\(r\)[/tex]:
[tex]\[ r^2 = \frac{135}{8} \times \frac{2}{15} = \frac{135}{60} = \frac{9}{4} \][/tex]
[tex]\[ r = \sqrt{\frac{9}{4}} = \frac{3}{2} \][/tex]
Thus, the general term [tex]\(g_n\)[/tex] is:
[tex]\[ g_n = a_1 \cdot r^{n-1} = \frac{15}{2} \cdot \left(\frac{3}{2}\right)^{n-1} \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{5\left(\frac{3}{2}\right)^{n-1}} \][/tex]
Problem 45:
What is the [tex]\(15^{\text{th}}\)[/tex] partial sum of the sequence [tex]\(-10, -1, 8, 17, \ldots\)[/tex]?
To find the partial sum, we first notice the pattern and identify whether it is an arithmetic sequence. We calculate the common difference [tex]\(d\)[/tex]:
[tex]\[ d = a_2 - a_1 = -1 - (-10) = 9 \][/tex]
[tex]\[ d = a_3 - a_2 = 8 - (-1) = 9 \][/tex]
Since this is an arithmetic sequence with:
- First term [tex]\(a = -10\)[/tex]
- Common difference [tex]\(d = 9\)[/tex]
The [tex]\(n^{\text{th}}\)[/tex] term of the sequence can be found by:
[tex]\[ a_n = a + (n-1)d = -10 + (n-1) \cdot 9 \][/tex]
The partial sum [tex]\(S_n\)[/tex] of the first [tex]\(n\)[/tex] terms of an arithmetic sequence is:
[tex]\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \][/tex]
For [tex]\(n = 15\)[/tex]:
[tex]\[ S_{15} = \frac{15}{2} \left(2(-10) + (15-1) \cdot 9\right) \][/tex]
[tex]\[ S_{15} = \frac{15}{2} \left(-20 + 126\right) \][/tex]
[tex]\[ S_{15} = \frac{15}{2} \cdot 106 = 15 \cdot 53 = 795 \][/tex]
Therefore, the [tex]\(15^{\text{th}}\)[/tex] partial sum is:
[tex]\[ \boxed{795} \][/tex]
Problem 46:
Which one of the following series is convergent?
Let's analyze each series:
A:
[tex]\[ \sum_{n=1}^{\infty}\left(\frac{\sqrt{5}}{2}\right)^n \][/tex]
This is a geometric series with [tex]\(a = \frac{\sqrt{5}}{2}\)[/tex] and [tex]\(r = \frac{\sqrt{5}}{2}\)[/tex]. Since [tex]\(\left|\frac{\sqrt{5}}{2}\right| > 1\)[/tex], the series diverges.
B:
[tex]\[ \sum_{n=1}^{\infty}(-1)^{n+1}\left(\frac{4}{5}\right)^n \][/tex]
This is an alternating series with [tex]\(|\text{ratio}| = \frac{4}{5} < 1\)[/tex]. It converges by the Alternating Series Test.
C:
[tex]\[ \sum_{n=1}^{\infty} 3^{n-3} \][/tex]
This can be rewritten as:
[tex]\[ \sum_{n=1}^{\infty} \frac{1}{27} \cdot 3^n \][/tex]
This is a geometric series with [tex]\(r = 3 > 1\)[/tex]. Therefore, it diverges.
D:
[tex]\[ \sum_{n=1}^{\infty}\left(\frac{4}{3}\right)^{n+1}\left(\frac{3}{2}\right)^{n-1} \][/tex]
Combining the terms, we get:
[tex]\[ \left(\frac{4}{3} \cdot \frac{3}{2}\right)^n \cdot \frac{4}{3} \left(\frac{3}{2}\right)^{-1} = \left(\frac{4}{2}\right)^n \cdot \left(\frac{4}{3} \cdot \frac{2}{3}\right) = 2^n \cdot \frac{8}{9} \][/tex]
Thus, it grows exponentially and diverges.
Therefore, the correct answer is:
[tex]\[ \boxed{\sum_{n=1}^{\infty}(-1)^{n+1}\left(\frac{4}{5}\right)^n} \][/tex]
Problem 47:
If the cost of a certain product in Birr as a function of time in days is given by [tex]\(f(t)=3t+t^2\)[/tex], then what is the average rate of change on the interval [tex]\(2 \leq t \leq 6\)[/tex]?
The average rate of change of a function [tex]\(f(t)\)[/tex] over an interval [tex]\([a, b]\)[/tex] is given by:
[tex]\[ \text{Average rate of change} = \frac{f(b) - f(a)}{b - a} \][/tex]
For [tex]\(f(t) = 3t + t^2\)[/tex], with [tex]\(a = 2\)[/tex] and [tex]\(b = 6\)[/tex]:
[tex]\[ f(6) = 3(6) + 6^2 = 18 + 36 = 54 \][/tex]
[tex]\[ f(2) = 3(2) + 2^2 = 6 + 4 = 10 \][/tex]
Therefore, the average rate of change is:
[tex]\[ \text{Average rate of change} = \frac{54 - 10}{6 - 2} = \frac{44}{4} = 11 \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{11} \][/tex]
Problem 43:
What is the [tex]\(6^{\text{th}}\)[/tex] term of a sequence whose general term is [tex]\(a_n = a_{n-1} + (n-1) a_{n-2}\)[/tex], where [tex]\(a_1 = 1\)[/tex] and [tex]\(a_2 = 3\)[/tex]?
Let's find the terms of the sequence step-by-step:
- [tex]\(a_1 = 1\)[/tex]
- [tex]\(a_2 = 3\)[/tex]
Now, using the given formula [tex]\(a_n = a_{n-1} + (n-1) a_{n-2}\)[/tex]:
- [tex]\(a_3 = a_2 + 2 \cdot a_1 = 3 + 2 \cdot 1 = 3 + 2 = 4\)[/tex]
- [tex]\(a_4 = a_3 + 3 \cdot a_2 = 4 + 3 \cdot 3 = 4 + 9 = 13\)[/tex]
- [tex]\(a_5 = a_4 + 4 \cdot a_3 = 13 + 4 \cdot 4 = 13 + 16 = 29\)[/tex]
- [tex]\(a_6 = a_5 + 5 \cdot a_4 = 29 + 5 \cdot 13 = 29 + 65 = 94\)[/tex]
Therefore, the sixth term of the sequence is:
[tex]\[ \boxed{94} \][/tex]
The correct answer is none of the given options, indicating possibly a miscalculation or error in listing the correct choices.
Problem 44:
What is the general term [tex]\(g_n\)[/tex] of the geometric sequence whose first term is [tex]\(\frac{15}{2}\)[/tex] and third term is [tex]\(\frac{135}{8}\)[/tex]?
Let's find the common ratio [tex]\(r\)[/tex]:
We know:
- [tex]\(a_1 = \frac{15}{2}\)[/tex]
- [tex]\(a_3 = \frac{135}{8}\)[/tex]
In a geometric sequence, [tex]\(a_3 = a_1 \cdot r^2\)[/tex]. Using the given values:
[tex]\[ \frac{135}{8} = \frac{15}{2} \cdot r^2 \][/tex]
Solving for [tex]\(r\)[/tex]:
[tex]\[ r^2 = \frac{135}{8} \times \frac{2}{15} = \frac{135}{60} = \frac{9}{4} \][/tex]
[tex]\[ r = \sqrt{\frac{9}{4}} = \frac{3}{2} \][/tex]
Thus, the general term [tex]\(g_n\)[/tex] is:
[tex]\[ g_n = a_1 \cdot r^{n-1} = \frac{15}{2} \cdot \left(\frac{3}{2}\right)^{n-1} \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{5\left(\frac{3}{2}\right)^{n-1}} \][/tex]
Problem 45:
What is the [tex]\(15^{\text{th}}\)[/tex] partial sum of the sequence [tex]\(-10, -1, 8, 17, \ldots\)[/tex]?
To find the partial sum, we first notice the pattern and identify whether it is an arithmetic sequence. We calculate the common difference [tex]\(d\)[/tex]:
[tex]\[ d = a_2 - a_1 = -1 - (-10) = 9 \][/tex]
[tex]\[ d = a_3 - a_2 = 8 - (-1) = 9 \][/tex]
Since this is an arithmetic sequence with:
- First term [tex]\(a = -10\)[/tex]
- Common difference [tex]\(d = 9\)[/tex]
The [tex]\(n^{\text{th}}\)[/tex] term of the sequence can be found by:
[tex]\[ a_n = a + (n-1)d = -10 + (n-1) \cdot 9 \][/tex]
The partial sum [tex]\(S_n\)[/tex] of the first [tex]\(n\)[/tex] terms of an arithmetic sequence is:
[tex]\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \][/tex]
For [tex]\(n = 15\)[/tex]:
[tex]\[ S_{15} = \frac{15}{2} \left(2(-10) + (15-1) \cdot 9\right) \][/tex]
[tex]\[ S_{15} = \frac{15}{2} \left(-20 + 126\right) \][/tex]
[tex]\[ S_{15} = \frac{15}{2} \cdot 106 = 15 \cdot 53 = 795 \][/tex]
Therefore, the [tex]\(15^{\text{th}}\)[/tex] partial sum is:
[tex]\[ \boxed{795} \][/tex]
Problem 46:
Which one of the following series is convergent?
Let's analyze each series:
A:
[tex]\[ \sum_{n=1}^{\infty}\left(\frac{\sqrt{5}}{2}\right)^n \][/tex]
This is a geometric series with [tex]\(a = \frac{\sqrt{5}}{2}\)[/tex] and [tex]\(r = \frac{\sqrt{5}}{2}\)[/tex]. Since [tex]\(\left|\frac{\sqrt{5}}{2}\right| > 1\)[/tex], the series diverges.
B:
[tex]\[ \sum_{n=1}^{\infty}(-1)^{n+1}\left(\frac{4}{5}\right)^n \][/tex]
This is an alternating series with [tex]\(|\text{ratio}| = \frac{4}{5} < 1\)[/tex]. It converges by the Alternating Series Test.
C:
[tex]\[ \sum_{n=1}^{\infty} 3^{n-3} \][/tex]
This can be rewritten as:
[tex]\[ \sum_{n=1}^{\infty} \frac{1}{27} \cdot 3^n \][/tex]
This is a geometric series with [tex]\(r = 3 > 1\)[/tex]. Therefore, it diverges.
D:
[tex]\[ \sum_{n=1}^{\infty}\left(\frac{4}{3}\right)^{n+1}\left(\frac{3}{2}\right)^{n-1} \][/tex]
Combining the terms, we get:
[tex]\[ \left(\frac{4}{3} \cdot \frac{3}{2}\right)^n \cdot \frac{4}{3} \left(\frac{3}{2}\right)^{-1} = \left(\frac{4}{2}\right)^n \cdot \left(\frac{4}{3} \cdot \frac{2}{3}\right) = 2^n \cdot \frac{8}{9} \][/tex]
Thus, it grows exponentially and diverges.
Therefore, the correct answer is:
[tex]\[ \boxed{\sum_{n=1}^{\infty}(-1)^{n+1}\left(\frac{4}{5}\right)^n} \][/tex]
Problem 47:
If the cost of a certain product in Birr as a function of time in days is given by [tex]\(f(t)=3t+t^2\)[/tex], then what is the average rate of change on the interval [tex]\(2 \leq t \leq 6\)[/tex]?
The average rate of change of a function [tex]\(f(t)\)[/tex] over an interval [tex]\([a, b]\)[/tex] is given by:
[tex]\[ \text{Average rate of change} = \frac{f(b) - f(a)}{b - a} \][/tex]
For [tex]\(f(t) = 3t + t^2\)[/tex], with [tex]\(a = 2\)[/tex] and [tex]\(b = 6\)[/tex]:
[tex]\[ f(6) = 3(6) + 6^2 = 18 + 36 = 54 \][/tex]
[tex]\[ f(2) = 3(2) + 2^2 = 6 + 4 = 10 \][/tex]
Therefore, the average rate of change is:
[tex]\[ \text{Average rate of change} = \frac{54 - 10}{6 - 2} = \frac{44}{4} = 11 \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{11} \][/tex]
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