Let's solve the problems shown in the image.

I. Find the first three terms of the sequence whose $n$-th term is given by the formula.

1. $a_n = n - 1$ $a_1 = 1 - 1 = 0, \quad a_2 = 2 - 1 = 1, \quad a_3 = 3 - 1 = 2$ So, the first three terms are: $0, 1, 2$.

2. $a_n = \frac{1}{n+1}$ $a_1 = \frac{1}{1+1} = \frac{1}{2}, \quad a_2 = \frac{1}{2+1} = \frac{1}{3}, \quad a_3 = \frac{1}{3+1} = \frac{1}{4}$ The first three terms are: $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$.

3. $a_n = \frac{4n}{n+2}$ $a_1 = \frac{4(1)}{1+2} = \frac{4}{3}, \quad a_2 = \frac{4(2)}{2+2} = \frac{8}{4} = 2, \quad a_3 = \frac{4(3)}{3+2} = \frac{12}{5}$ The first three terms are: $\frac{4}{3}, 2, \frac{12}{5}$.

4. $a_n = 3n^2 + 4$ $a_1 = 3(1)^2 + 4 = 3 + 4 = 7, \quad a_2 = 3(2)^2 + 4 = 12 + 4 = 16, \quad a_3 = 3(3)^2 + 4 = 27 + 4 = 31$ The first three terms are: $7, 16, 31$.

5. $a_n = 2n - 3$ $a_1 = 2(1) - 3 = 2 - 3 = -1, \quad a_2 = 2(2) - 3 = 4 - 3 = 1, \quad a_3 = 2(3) - 3 = 6 - 3 = 3$ The first three terms are: $-1, 1, 3$.

II. Use the difference table to predict the next term in the sequence. Show your complete formula.

1. $1, 6, 22, 56, 114, 202$ The differences between terms are: $6 - 1 = 5, \quad 22 - 6 = 16, \quad 56 - 22 = 34, \quad 114 - 56 = 58, \quad 202 - 114 = 88$ The second differences: $16 - 5 = 11, \quad 34 - 16 = 18, \quad 58 - 34 = 24, \quad 88 - 58 = 30$ Third differences are constant: $18 - 11 = 7, \quad 24 - 18 = 6, \quad 30 - 24 = 6$ Since the third differences are nearly constant, this sequence is cubic. The next term can be predicted by continuing the third differences, leading to the next term: 326.

2. $9, 15, 21, 27, \dots$ The differences are constant: $15 - 9 = 6, \quad 21 - 15 = 6, \quad 27 - 21 = 6$ This is an arithmetic sequence with a common difference of 6. The next term is $27 + 6 = 33$.

3. $2, 6, 12, 20, 30, 42, \dots$ The differences are: $6 - 2 = 4, \quad 12 - 6 = 6, \quad 20 - 12 = 8, \quad 30 - 20 = 10, \quad 42 - 30 = 12$ The differences increase by 2, so the next difference will be 14, leading to the next term: $42 + 14 = 56$.

4. $6, 11, 18, 27, 38, \dots$ The differences are: $11 - 6 = 5, \quad 18 - 11 = 7, \quad 27 - 18 = 9, \quad 38 - 27 = 11$ The differences increase by 2, so the next difference will be 13, leading to the next term: $38 + 13 = 51$.

5. $1, 8, 21, 40, 65, 96, \dots$ The differences are: $8 - 1 = 7, \quad 21 - 8 = 13, \quad 40 - 21 = 19, \quad 65 - 40 = 25, \quad 96 - 65 = 31$ The differences increase by 6, so the next difference will be 37, leading to the next term: $96 + 37 = 133$.

Let me know if you'd like further details or have any questions!

Related Questions:

1. How do you determine if a sequence is arithmetic or geometric?
2. What are difference tables used for in sequences?
3. How do you find the formula for a quadratic sequence?
4. What is the general form of a cubic sequence formula?
5. How can you use the first differences to predict the next term in a sequence?

Tip:

For sequences that exhibit constant second or third differences, the sequence can often be modeled by a quadratic or cubic function respectively.