Question

Part A

Directions: Try to find the first three numbers in the series, using the given rule.

1. tn = 2(n + 3)

2. tn = 3n – 1

3. tn = n + 11

4. tn = 5(2)^n

5. tn = (-2)^n

6. tn = n^n

7. tn = n^nn + 1

8. tn = n(n^n – 1)

9. tn = 2^n + 1

10. tn = 1/n

Answer 1

Sure, here are the first three numbers in each series:

1. tn = 2(n + 3)
- For n = 1: t1 = 2(1 + 3) = 8
- For n = 2: t2 = 2(2 + 3) = 10
- For n = 3: t3 = 2(3 + 3) = 12

2. tn = 3n – 1
- For n = 1: t1 = 3(1) - 1 = 2
- For n = 2: t2 = 3(2) - 1 = 5
- For n = 3: t3 = 3(3) - 1 = 8

3. tn = n + 11
- For n = 1: t1 = 1 + 11 = 12
- For n = 2: t2 = 2 + 11 = 13
- For n = 3: t3 = 3 + 11 = 14
4. tn = 5(2)^n
- For n = 1: t1 = 5(2)^1 = 10
- For n = 2: t2 = 5(2)^2 = 20
- For n = 3: t3 = 5(2)^3 = 40

5. tn = (-2)^n
- For n = 1: t1 = (-2)^1 = -2
- For n = 2: t2 = (-2)^2 = 4
- For n = 3: t3 = (-2)^3 = -8

6. tn = n^n
- For n = 1: t1 = 1^1 = 1
- For n = 2: t2 = 2^2 = 4
- For n = 3: t3 = 3^3 = 27

7. tn = n^nn + 1
- For n = 1: t1 = 1^2 = 1
- For n = 2: t2 = 2^3 = 8
- For n = 3: t3 = 3^4 = 81

8. tn = n(n^n – 1)
- For n = 1: t1 = 1(1^1 - 1) = 0
- For n = 2: t2 = 2(2^2 - 1) = 6
- For n = 3: t3 = 3(3^3 - 1) = 78

9. tn = 2^n + 1
- For n = 1: t1 = 2^1 + 1 = 3
- For n = 2: t2 = 2^2 + 1 = 5
- For n = 3: t3 = 2^3 + 1 = 9

10. tn = 1/n
- For n = 1: t1 = 1/1 = 1
- For n = 2: t2 = 1/2 ≈ 0.5
- For n = 3: t3 = 1/3 ≈ 0.333

Let me know if you need further assistance!