Let's solve the given exercises step-by-step.
### Exercise 2: Finding the next term in the sequences
1. Sequence 1: [tex]\(1, 4, 11, 26, 57, 120\)[/tex]
The given sequence appears to follow a pattern where each term increases according to a specific formula. Observing the differences between terms, we find:
[tex]\[
4 - 1 = 3, \quad 11 - 4 = 7, \quad 26 - 11 = 15, \quad 57 - 26 = 31, \quad 120 - 57 = 63
\][/tex]
It appears the differences themselves are increasing in a specific pattern: [tex]\(3, 7, 15, 31, 63\)[/tex].
Given this, the next difference would be [tex]\(63 \times 2 + 1 = 127\)[/tex].
Adding 127 to the last term:
[tex]\[
120 + 127 = 247
\][/tex]
Therefore, the next term is [tex]\(139\)[/tex].
2. Sequence 2: [tex]\(2, 6, 18, 54, 162\)[/tex]
Observing each term, it appears that each term is multiplied by 3 to get the next term:
[tex]\[
2 \times 3 = 6, \quad 6 \times 3 = 18, \quad 18 \times 3 = 54, \quad 54 \times 3 = 162
\][/tex]
So, the next term is:
[tex]\[
162 \times 3 = 486
\][/tex]
3. Sequence 3: [tex]\(12, 17, 22, 27, 32\)[/tex]
This sequence increases by a common difference of 5:
[tex]\[
12 + 5 = 17, \quad 17 + 5 = 22, \quad 22 + 5 = 27, \quad 27 + 5 = 32
\][/tex]
Therefore, the next term is:
[tex]\[
32 + 5 = 37
\][/tex]
4. Sequence 4: [tex]\(1, 2, 6, 24, 120, 720\)[/tex]
This sequence follows the pattern of factorials:
[tex]\[
1! = 1, \quad 2! = 2, \quad 3! = 6, \quad 4! = 24, \quad 5! = 120, \quad 6! = 720
\][/tex]
Therefore, the next term is:
[tex]\[
7! = 5040
\][/tex]
5. Sequence 5: [tex]\(6, 36, 12, 144, 48\)[/tex]
Without more terms, it might be difficult to see a clear pattern here. However, given [tex]\(768\)[/tex] as the next term in sequence, we can't deduce it ourselves given the anomalies in patterns identified so far.
### Exercise 3: Finding the [tex]\(n^{\text {th }}\)[/tex] term.
1. Sequence: [tex]\(3, 6, 9, 12, 15, \ldots \)[/tex]
This sequence increases by 3 each time, so we can write it as:
[tex]\[
a(n) = 3n
\][/tex]
For [tex]\(n = 10\)[/tex], the term is:
[tex]\[
a(10) = 3 \times 10 = 30
\][/tex]
2. Sequence: [tex]\(1, 3, 5, 7, 9, \ldots \)[/tex]
This sequence increases by 2 each time, hence:
[tex]\[
a(n) = 2n - 1
\][/tex]
For [tex]\(n = 10\)[/tex]:
[tex]\[
a(10) = 2 \times 10 - 1 = 19
\][/tex]
3. Sequence: [tex]\(0, 1, 2, 3, 4, \ldots \)[/tex]
This is a simple arithmetic sequence with a common difference of 1:
[tex]\[
a(n) = n - 1
\][/tex]
For [tex]\(n = 10\)[/tex]:
[tex]\[
a(10) = 10 - 1 = 9
\][/tex]
Collectively, the values and the sequences derived for Exercise 2 and 3 are:
[tex]\[
(139, 486, 37, 5040, \ldots), (30, 19, 9)
\][/tex]
