1.Find a recurrence relation and initial conditions for sequences: 6,
13, 27, 55.
2.Write the first 4 terms in the sequence an+1 = 3an
2 -9 where n≥1 and a1=2.
sure looks like a_n+1 = 2a_n + 1
really? You can't just plug in values for n?
If 3an
2
means 3an/2 then
a2 = 3a1/2-9 = 3*1-9 = -6
If not, you'll have to clean it up on your own
1. To find a recurrence relation for the sequence 6, 13, 27, 55, we need to identify the pattern or relationship between the terms.
- Let's start by examining the differences between consecutive terms:
- 13 - 6 = 7
- 27 - 13 = 14
- 55 - 27 = 28
- Now, let's examine the differences between these differences:
- 14 - 7 = 7
- 28 - 14 = 14
- Notice that the second differences are constant. This suggests that the sequence might be an arithmetic sequence with a quadratic relationship.
- To confirm this, let's divide the differences by the corresponding squares of the counting numbers:
- 7 / 1^2 = 7
- 14 / 2^2 = 7
- As both divisions give the same result, our assumption is correct.
- This means that the sequence follows the formula: an = an-1 + dn + cn^2, where d is the common difference between consecutive terms, and c is a constant.
- Let's assign variables to the differences and represent the given sequence using this formula. Since the first term is given explicitly, a0 will be the constant term:
- a1 = 6 = a0 + d + c(0)^2 = a0 + d
- a2 = 13 = a1 + d + c(1)^2 = (a0 + d) + d + c
- a3 = 27 = a2 + d + c(2)^2 = (a0 + d + d + c) + d + 4c
- a4 = 55 = a3 + d + c(3)^2 = (a0 + d + d + c + d + 4c) + d + 9c
- Simplifying each equation, we get:
- a1 = a0 + d
- a2 = a0 + 2d + c
- a3 = a0 + 3d + 5c
- a4 = a0 + 4d + 13c
- Now, let's find the recurrence relation using these equations:
- a1 - a0 = d => d = a1 - a0
- a2 - a1 = 2d + c => 2d + c = (a2 - a1)
- a3 - a2 = 3d + 5c => 3d + 5c = (a3 - a2)
- a4 - a3 = 4d + 13c => 4d + 13c = (a4 - a3)
- Substituting the values, we have:
- d = 13 - 6 = 7
- (2d + c) = 27 - 13 = 14
- (3d + 5c) = 55 - 27 = 28
- (4d + 13c) = ??? (We need another term to evaluate this)
- You can now solve these equations to find the values of d and c, or you can use other terms in the sequence to find the fourth equation to solve for (4d + 13c).
2. The given sequence is defined by the recurrence relation an+1 = 3an - 9, where n ≥ 1 and a1 = 2.
- To find the first four terms, we can use the recurrence relation and the given initial condition:
- a1 = 2 (given)
- For n = 2: a2 = 3a1 - 9 = 3*2 - 9 = -3
- For n = 3: a3 = 3a2 - 9 = 3*(-3) - 9 = -18
- For n = 4: a4 = 3a3 - 9 = 3*(-18) - 9 = -63
So, the first four terms in the sequence are 2, -3, -18, -63.